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Solution of hydrogen in palladium is such an example in which solute is a gas and solvent is solid. Mole fraction. Mole fraction is defined as the ratio of number of moles of a component and total number of moles in all components.
It is defined as the number of moles of solute dissolved per kg g of solvent. It is independent of temperature. Molarity is defined as number of moles of solute dissolved per litre or ml of solution. It depends on temperature because volume is dependent on temperature. Mass percentage. Mass percentage is defined as the percentage ratio of mass of one component to the total mass of all the components.
What should be the molarity of such a sample of the acid if the density of the solution is. According to given question, in g of solution 68 g is nitric acid and rest is water. So moles of 68 g HNO 3 Density of solution is given to be 1. So volume of g solution becomes Thus, molarity of nitric acid is :. If the density of solution is then what shall be the molarity of the solution?
According to question, mass percentage means in g of solution 10 g glucose is dissolved in 90 g water. So moles of glucose are :. Mole fraction Molarity :- Volume of g solution :. Let the amount of Na 2 CO 3 be x g. So the amount of NaHCO 3 will be equal to 1 - x g. Now it is given that it is an equimolar mixture. It is given that molarity of HCl is 0.
Thus required volume :. Calculate the mass percentage of the resulting solution. According to question we have 2 solute,. Solute 1. Solute 2. Thus mass percentage of solute is :. Calculate the molality of the solution. If the density of the solution is then what shall be the molarity of the solution? For finding molality we need to find the moles of ethylene glycol.
Moles of ethylene glycol :. Now for molarity Volume of solution. So molarity The level of contamination was by mass :. We know that 15 ppm means 15 parts per million. Required percent by mass :. Moles of chloroform :. Mass of water is. Since contamination is 15 ppm. So molality will be :. Both alcohol and water individually have strong hydrogen bonds as their force of attraction.
When we mix alcohol with water they form solution due to the formation of hydrogen bonds but they are weaker as compared to hydrogen bonds of pure water or pure alcohol. Thus this solution shows a positive deviation from the ideal behaviour. It is known that dissolution of gas in a liquid is an exothermic process.
So, by Le Chatelier principle we know that equilibrium shifts backwards as we increase temperature in case of exothermic process. Thus gases always tend to be less soluble in liquids as the temperature is raised. According to Henry's law at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of solution or liquid. Some of its applications are as follows This leads to low concentrations of oxygen in the blood and tissues climbers.
Due to low blood oxygen, climbers become weak and unable to think clearly which are symptoms of a condition known as anoxia. If the solution contains of ethane, then what shall be the partial pressure of the gas? Using Henry's Law we can write,. Putting value in this equation, we get :. So, the magnitude of k is. Now, we will again use the above equation for. So the required partial pressure is Enthalpy relation to positive and negative deviation can be understood from the following example Consider a solution made up of two components - A and B.
In the pure state the intermolecular force of attraction between them are A-A and B-B. But when we mix the two, we get a binary solution with molecular interaction A-B. If A-B interaction is weak than A-A and B-B then enthalpy of reaction will be positive thus reaction will tend to move in a backward direction. Hence molecules in binary solution will have a higher tendency to escape. Thus vapour pressure increases and shows positive deviation from the ideal behaviour.
What is the molar mass of the solute? It is given that of aq. This means 2 g of non-volatile solute in 98 g of H 2 O. Using Raoult's law :. Thus the molar mass of non-volatile solute is At , the vapour pressures of the two liquid components are and respectively. What will be the vapour pressure of a mixture of of heptane and of octane? Firstly we will find moles of heptane and octane so that we can find vapour pressure of each. So moles of heptane :.
Now we will find the partial vapour pressure It is asked the vapour pressure of 1 molal solution which means 1 mol of solute in g H 2 O. Since the molecular weight of H 2 O is Mole fraction of solute :.
Applying the equation :. Thus the vapour pressure of the solution is After adding solute to octane, the vapour pressure becomes :. Moles of octane :. Using Raoult's law we get :. Further, of water is then added to the solution and the new vapour pressure becomes at. In this question we will find molar mass of solute by using Raoult's law. Let the molar mass of solute is M. Initially we have 30 g solute and 90 g water. By Raoult's law we have Now we have added 18 g of water more, so the equation becomes:.
Putting this in above equation we obtain From equation i and ii we get. So the molar mass of solute is 23 units. In the previous part we have calculated the value of molar mass the Raoul's law equation. We had It is given that freezing point of pure water is Moles of cane sugar :. Molality :. We also know that -. Now we will use the above procedure for glucose. Moles of glucose :. Thus molality :. So, we can find the elevation in freezing point:. Thus freezing point of glucose solution is When dissolved in 20 g of benzene of AB 2 lowers the freezing point by whereas of AB 4 lowers it by.
The molar depression constant for benzene is Calculate atomic masses of A and B. In this question we will use the formula :. Firstly for compound AB 2 Similarly for compound AB 4 If we assume atomic weight of element A to be x and of element B to be y, then we have Solving both the equations, we get Hence atomic mass of element A is According to given conditions we have same solution under same temperature.
So we can write :. So, if we put all the given values in above equation, we get. Hence the required concentration is 0. Since both the compounds are alkanes so their mixture has van der Waal force of attraction between compounds.
The binary mixture of these compounds has van der Waal force of attraction between them. The given compounds will have ion-dipole interaction between them. Methanol has -OH group and acetone has ketone group. So there will be hydrogen bonding between them. They will have dipole-dipole interaction since both are polar compounds. In this, we have used the fact that like dissolves like. Since cyclohexane is an alkane so its solubility will be maximum.
We know the fact that like dissolves in like. Since phenol is had both polar and non-polar group so it is partially soluble in water. Since toluene is a non-polar compound i. Since the -OH group in formic acid polar can form H-bonds with water thus it is highly soluble in water. Ethylene glycol is an organic compound but is polar in nature. Also, it can form H-bonds with water molecules, thus it is highly soluble in water.
Chloroform is a non-polar compound so it is insoluble in water. Pentanol has both polar and non-polar groups so it is partially soluble in water. We know that, Molality :. So, for moles of solute we have :. Thus, molality :. We are given,. The dissociation equation of CuS is given by So, the equation becomes Thus maximum molarity of solution is.
Thus the mass percentage of aspirin is. Dose of nalorphene generally given is Calculate the mass of aqueous solution required for the above dose.
We are given with molality of the solution, so we need to find the moles of Nalorphene. So moles of nalorphene :. So the required weight of water is 3. We are given with the molarity of solution. So mass of benzoic acid :. Hence the required amount of benzoic acid is 4. Explain briefly. We know that depression in freezing point of water will depend upon the degree of ionisation. The degree of ionisation will be highest in the case of trifluoroacetic acid as it is most acidic among all three.
So the depression in freezing point will be reverse of the above order. Firstly we will find the Vant's Hoff factor the dissociation of given compound. So we can write,. Putting values of K a and C in the last result, we get :.
Thus, molality of the solution :. Now we will use :. The depression in the freezing point of water observed is. Firstly we need to calculate molality in order to get vant's hoff factor. We need to assume volume of solution to be nearly equal to mL. Now, we know that :. Now for dissociation constant Put values of C and a in the above equation, we get :. Firstly we will find number of moles of both water and glucose. Calculate the solubility of methane in benzene at under.
We are given value of P and k, so C can be found. Hence solubility of methane in benzene is. The vapour pressure of pure liquid B was found to be Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure the solution is.
For calculating partial vapour pressure we need to calculate mole fractions of components. So number of moles of liquid A :. Mole fraction of A x A :. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone.
The experimental data observed for different compositions of mixture is:. Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution. The vapour pressure of pure benzene and toluene at are and respectively.
Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with toluene. Firstly, we will find the no. Now we will find mol fraction of both Mole fraction of benzene Hence mole fraction of benzene in vapour phase is given by :. The major components are oxygen and nitrogen with approximate proportion of is to by volume at. The water is in equilibrium with air at a pressure of At. We have been given that the water is in equilibrium with air at a pressure of 10 atm or mm of Hg.
So the partial pressure of oxygen :. Now, by Henry's Law :. For oxygen :. For nitrogen :. Hence the mole fraction of nitrogen and oxygen in water is and respectively. We know that osmotic pressure :. We have been given the values of osmotic pressure, V, i and T. So the value of w can be found. Hence 3. Dissociation of K 2 SO 4 is as follows It is clear that 3 ions are produced, so the value of i will be 3. Putting all the values NCERT solutions for class 12 chemistry. NCERT solutions for class 12 subject wise.
Hope you have understood well with the help of the free solutions provided here. After completing NCERT solutions for class 12 chemistry chapter 2 Solutions, students will be able to differentiate between the types of solutions characteristics of ideal and non-ideal solutions, define solubility and colligative properties, understand abnormal molar mass. The solutions which you read here will also help you in building your concepts as well as a strong base in the subject.
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Updated on Aug 5, - p. IST by Sumit Saini. Know More Latest : Why settle for a classroom full of children when you can have your own personalized classroom? Know More Share. Table of contents. Start Now. Answer : We know that solute and solvent forms solution. So mass percentage of benzene solute :- Similarly mass percentage of CCl 4 The attractive force which holds various constituents atoms, ions, etc.
Different theories and concepts have been put forward from time to time to analyze the formation of the bond. Atoms, therefore combine with each other and complete their respective octets or duplets to attain stable configuration of the nearest noble gases. As it was seen that the noble gases are very stable and were inert to react to others. So, there is a sharing of electrons or transferring one or more electrons from one atom to another, as a result, a chemical bond is formed, known as a covalent bond or ionic bond.
The Lewis dot symbol of Mg atom is;. As there are two valence electrons in Mg atom. Hence, the Lewis dot symbol for Mg is:. The Lewis dot symbol of Na atom is;. As there is only one valence electron in Na atom of Na. Hence, the Lewis dot structure is. The Lewis dot symbol of atom is;. As there are three valence electrons in atom. As there are six valence electrons in an atom of. As there are five valence electrons in an atom of.
As there are seven valence electrons in an atom of. As the number of valence electrons in sulphur is six. Therefore its Lewis dot symbol of sulphur S is. And of is, if it has two electrons more because of its dinegative charge. As the number of valence electrons in aluminium is three.
Therefore its Lewis dot symbol of aluminium Al is. And of is, if it has donated three electrons because of its tripositive charge. As the number of valence electrons in hydrogen is one. Therefore its Lewis dot symbol of hydrogen H is. And of is, if it has one more electron because of its a negative charge develops. The Lewis structure of is:.
Write its significance and limitations. Atoms can combine either by transfer of valence electrons from one atom to another gaining or losing or by sharing of valence electrons in order to have an octet in their valence shells. This is known as the octet rule. Significance : It is quite useful for understanding the structures of most of the organic compounds and it applies mainly to the second-period elements of the periodic table.
Limitations : There are three types of exceptions to the octet rule. Formation of ionic bond takes place by the transfer of one or more electrons from one atom to another. So, ionic bond formation mainly depends upon the ease with which neutral atoms can lose or gain electrons. The bond formation also depends upon the lattice energy of the compound formed. Ionic bonds will be formed more easily between elements with comparatively low ionization enthalpies and elements with a comparatively high negative value of electron gain enthalpy.
THe central atom has no lone pair and there are two bond pairs. The central atom has no lone pair and there are three bond pairs. The central atom has no lone pair and there are four bond pairs. THe central atom has no lone pair and there are five bond pairs. Hence, it has trigonal bipyramidal shape. The central atom has no lone pair and there are two bond pairs. Hence, it has a bent shape. THe central atom has no lone pair and there are three bond pairs. Bond Strength gives us that amount of energy needed to break a bond between atoms forming a molecule.
So, with an increase in bond order, bond enthalpy increases as a result bond strength increases. Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. The bond length in a covalent molecule AB. The single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atoms is inadequate to represent the molecule accurately as it represents unequal bonds.
According to the experimental findings, all carbon to oxygen bonds in CO3 2� are equivalent. Therefore the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing.?
If not, give reasons for the same. As per the rule, it is not having the same position as the atoms and is changed. Hence the given structures cannot be taken as the canonical forms of the resonance hybrid. The resonance structures. K and S:. We have the electronic configurations of both:. So, there will be an electron transfer between them as follows:. So, there will be electron transfer between them as follows:. Explain this on the basis of dipole moment.
H2O molecule, which has a bent structure, the two O�H bonds are oriented at an angle of Net dipole moment of 6. While on the other hand. The dipole moment of carbon dioxide is zero. This may be because of linear shape of the molecule as it has two C-O bonds which has opposite dipole moments cancelling each other.
Some of the important significance of the dipole moment is as follows:. We can determine the shape of the molecule. Symmetrical molecules like linear, etc. For determining the polarity of the molecules.
Greater the dipole moment value, more will be the polarity and vice-versa. We can say that if a molecule has zero dipole moment then it must be non-polar and if it is non-zero then it must have some polar character. How does it differ from electron gain enthalpy?
Electronegativity is the ability of an atom in a compound to attract a bond pair of electrons towards itself. It cannot be measured and it is a relative number. The electron gain enthalpy, , is the enthalpy change, when a gas phase atom in its ground state gains an electron. The electron gain process may be exothermic or endothermic.
An element has a constant value of the electron gain enthalpy that can be measured experimentally. A polar covalent bond, when two different atoms are linked to each other by covalent bond, then the shared electron pair will not be in the centre just because the bonding atoms differ in electronegativities.
For examples, in ,. Here slightly positive charges are developed in hydrogen atoms and slightly negative charge developed in oxygen atom as oxygen is more electronegative than the hydrogen. Thus, opposite poles are developed in the molecule.
Hence the bond pair lies towards oxygen atom. The ionic character in a molecule depends on the electronegativity difference between the constituting atoms.
More the difference more will be the ionic character of the molecule. So, on this basis, we have the order of increasing ionic character in the given molecules. Here hydrogen atom is bonded to carbon with a double bond, which is not possible because hydrogen has only one electron to share with carbon. Also, the second carbon does not have its valency satisfied. Therefore, the correct skeletal structure of as shown below:. Explain why is not square planar?
The electronic configuration of carbon atom is. Where it has s-orbital, p-orbital only and there is no d-orbital present. Hence the carbon atom undergoes hybridization in methane molecule and takes a tetrahedral shape. And for a molecule to have a square planar structure it must have d orbital present.
But here the absence of d-orbital, as a result, it does not undergo hybridization, the structure of methane cannot be square planar. Also the reason that bond angle in square planar which makes the molecule more unstable because of repulsion between the bond pairs.
Here both have central atom Nitrogen and it has a lone pair of electrons with three bond pairs. Hence both molecules have a pyramidal shape. The electronegativity of fluorine is more as compared to the hydrogen.
Hence it is expected that the net dipole moment of is greater than. However has the net dipole moment of 1. This is because of the direction of the dipole moments of each individual bond in and. The moments of the lone pair in partly cancel out. But in the resultant moment add up to the bond moment of the lone pair. Describe the shapes of , , hybrid orbitals. Hybridisation which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.
The shapes of , , hybrid orbitals are shown:. Initially, the aluminium is in the ground state and the valence orbital can be shown as:. Then the electron gets excited so, the valence orbital can be shown as:. So, initially, aluminium was hybridisation and hence having a trigonal planar shape. Then it reacts with chloride ion to form. Where it has the empty orbital which gets involved and the hybridisation changes from. Hence there is a shape change from trigonal planar to tetrahedral.
Initially boron atom was in hybridised. The valence orbital of boron in the excited state can be shown as:. And nitrogen atom in is hybridised. The valence orbital of nitrogen in the excited state can be shown as:. Then after the reaction has occured the product is formed by the hybridisation of 'B' changes to. However, the hybridisation of 'N' remains unchanged. We have the electronic configuration of C-atom in the excited state is:.
Formation of an ethane molecule by overlapping of a hybridized orbital of another carbon atom, thereby forming a sigma bond. The remaining two orbitals of each carbon atom from a sigma bond with two hydrogen atoms. The unhybridized orbital of one carbon atom undergoes sidewise overlap with the orbital of a similar kind present on another carbon atom to form a weak n-bond.
Formation of molecule, each C-atom is sp hybridized with two 2p-orbitals in an unhybridized state. So the triple bond between the two carbon atoms is made up of one sigma and two pi bonds as shown in Fig. Given molecule :. Orbitals will form a sigma bond as both orbitals are spherical and can combine along x-axis as the internuclear axis.
Orbitals will form a sigma bond as 1s orbital and 2p x orbital are align such that they can combine along x-axis as the internuclear axis. Orbitals will not form a sigma bond as both 2p y orbital are align in y -direction but the internuclear axis is x-axis. Formation of pi bond takes place.
Orbitals will form a sigma bond as both 1s and 2s orbitals are spherical and can combine along the x-axis as the internuclear axis. There are 4 sigma bonds single bond each with the help of one s hybrid orbital and 3 p hybrid orbital, Hence C 1 and C 2 are hybridized.
While are making a double bond. Therefore they both are hybridized. Illustrate by giving one exmaple of each type. The shared pairs of electrons present between the bonded atoms are called bond pairs. And all valence electrons may not participate in bonding that electron pairs that do not participate in bonding are called lone pairs of electrons.
For examples,. In ethane, there are seven bond pairs but no lone pair is present. In , there are two bond pairs and two lone pairs on the central atom oxygen. Difference between the sigma bond and the pi bond is shown in the table below:. Formation of molecule:. Assume that two hydrogen atoms with nuclei and electrons are taken to undergo a reaction to form a hydrogen molecule.
When the two atoms are at a large distance, there is no interaction between them. As they approach each other, the attractive and repulsive forces start operating. Attractive force arises between:. Repulsive force arises between:. The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart. The attractive force overcomes the repulsive force. Hence, the two atoms approach each other. As a result, the potential energy decreases.
Finally, a state is achieved when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule. The important conditions required for the linear combination of atomic orbitals to form molecular orbitals are as follows:. The combining atomic orbitals must have the same or nearly the same energy.
The combining atomic orbitals must have the same symmetry about the molecular axis. The combining atomic orbitals must overlap to the maximum extent. The electronic configuration of Be is. From the molecular orbital electronic configuration, we have for molecule,. We can calculate the bond order for is where,. So, therefore we have,.
Bond order of. Hence, molecule does not exist. The electronic configuration of molecule can be written as:. Here the number of bonding electrons is and the number of antibonding electrons is. Therefore, the bond dissociation energy is directly proportional to the bond order.
Thus, the higher the bond order, the greater will be the stability. We get this order of stability:. Wave functions can be used to represent molecular orbitals. The plus and minus represent the positive wave function while negative wave function respectively. The initial ground state and final excited state electronic configuration of phosphorus P are:.
So, the phosphorus atom is hybridized in the excited state. The donated electron pairs by five Cl atoms are filled and make. The resultant shape is trigonal bipyramidal and the five hybrid orbitals are directed towards the five corners.
The five P-Cl sigma bonds, three lies in one plane and make with each other are equatorial bonds and the two P-Cl bonds lie above and below the equatorial plane makes an angle of with the plane are axial bonds. So, just because of more repulsion from the equatorial bond pairs, the axial bonds are slightly longer than equatorial bonds.
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