MODEL ANSWER PRACTICE TEST NO. 01CHEMISTRYHere, p = 760 mm HgkH = 4.27 × 105 mm Hg According to Henry’s law,p = kHx = 177.99 × 10−5= 178 × 10−5 (approximately)Hence, the mole fraction of methane in benzene is 178 × 10−5.FiveSimilar to liquids, glass has a tendency to flow, though very slowly. Therefore, glass is considered as a super cooled liquid. This is the reason that glass windows and doors are slightly thicker at the bottom than at the top.It is given that: KH = 1.67 × 108 Pa= 2.5 atm = 2.5 × 1.01325 × 105 Pa= 2.533125 × 105 PaAccording to Henry’s law:= 0.00152We can write, [Since, is negligible as compared to]In 500 mL of soda water, the volume of water = 500 mL[Neglecting the amount of soda present]We can write:500 mL of water = 500 g of water= 27.78 mol of waterNow, Hence, quantity of CO2 in 500 mL of soda water = (0.042 × 44)g= 1.848 gNumber of close-packed particles = 0.5 × 6.022 × 1023 = 3.011 × 1023Therefore, number of octahedral voids = 3.011 × 1023And, number of tetrahedral voids = 2 × 3.011 × 1023 = 6.022 ×1023Therefore, total number of voids = 3.011 × 1023 + 6.022 × 1023 = 9.033 × 1023The ccp lattice is formed by the atoms of the element N.Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N.According to the question, the atoms of element M occupy of the tetrahedral voids.Therefore, the number of atoms of M is equal to of the number of atoms of N.Therefore, ratio of the number of atoms of M to that of N is M: N Thus, the formula of the compound is M2 N3.(i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor.(ii) B (a group 13 element) is doped with Si (a group 14 element). So, there will be an extra electron and the semiconductor generated will be an n-type semiconductor.Mass percentage of C6H6 Mass percentage of CCl4Alternatively,Mass percentage of CCl4 = (100 − 15.28)%= 84.72%It is given that the density of niobium, d = 8.55 g cm−3Atomic mass, M = 93 gmol−1As the lattice is bcc type, the number of atoms per unit cell, z = 2We also know that, NA = 6.022 × 1023 mol−1Applying the relation:= 3.612 × 10−23 cm3So, a = 3.306 × 10−8 cmFor body-centred cubic unit cell:= 1.432 × 10−8 cm= 14.32 × 10−9 cm= 14.32 nmIt is given that vapour pressure of water, = 23.8 mm of HgWeight of water taken, w1 = 850 gWeight of urea taken, w2 = 50 gMolecular weight of water, M1 = 18 g mol−1Molecular weight of urea, M2 = 60 g mol−1Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1.Now, from Raoult’s law, we have:Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173(i) For cubic close-packed structure:= 353.55 pm= 354 pm (approximately)(ii) Volume of one unit cell = (354 pm)3= 4.4 × 107 pm3= 4.4 × 107 × 10−30 cm3= 4.4 × 10−23 cm3Therefore, number of unit cells in 1.00 cm3 = = 2.27 × 1022Or//(i) Simple cubicIn a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.Let the edge length of the cube be ‘a’ and the radius of each particle be r.So, we can write:a = 2rNow, volume of the cubic unit cell = a3= (2r)3= 8r3We know that the number of particles per unit cell is 1.Therefore, volume of the occupied unit cell Hence, packing efficiency (ii) Body-centred cubicIt can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.From ΔFED, we have:Again, from ΔAFD, we have:Let the radius of the atom be r.Length of the body diagonal, c = 4πor, Volume of the cube, A body-centred cubic lattice contains 2 atoms.So, volume of the occupied cubic lattice (iii) Face-centred cubicLet the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.From ΔABC, we have:Let r be the radius of the atom.Now, from the figure, it can be observed that:Now, volume of the cube, We know that the number of atoms per unit cell is 4.So, volume of the occupied unit cell = 74%A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that ΔPOQ is right-angled ∠POQ = 900Now, applying Pythagoras theorem, we can write:Percentage of oxygen (O2) in air = 20 %Percentage of nitrogen (N2) in air = 79%Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, (10 × 760) mm Hg = 7600 mm HgTherefore, Partial pressure of oxygen, = 1520 mm HgPartial pressure of nitrogen, = 6004 mmHgNow, according to Henry’s law:p = KH.xFor oxygen:For nitrogen:Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10−5and 9.22 × 10−5 respectively.Or//Molar mass of ∴No. of moles present in 10 g of It is given that 10 g of is added to 250 g of water.∴Molality of the solution, Let α be the degree of dissociation of undergoes dissociation according to the following equation:Since α is very small with respect to 1, 1 − α ≈ 1Now, Again,Total moles of equilibrium = 1 − α + α + α= 1 + αHence, the depression in the freezing point of water is given as: