Tiêu đề DPP-1 SOLUTION.pdf ✅

CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 1 1 (a) 1 atm = 76 cm = 76 × 13.6 × 980 dyne cm2 2 (a) Number of moles of He = 0.4 4 = 0.1 Number of moles of oxygen = 1.6 32 = 0.05 Number of moles of nitrogen = 1.4 28 = 0.05 Total moles in the 10.0 L cylinder at 27°C = 0.1 + 0.05 + 0.05 = 0.2 mol pT = nRT V = 0.2 × 0.082 × 300 10 = 0.492 atm 4 (c) At constant P,V and T,w ∝ m. 5 (b) In face centred cubic structure, contribution of 1 8 by each atompresent on the corner and 1 2 by each atom present on the face 6 (a) Rate of diffusion for H2 is maximum. 7 (d) Schottky defects occurs in highly ionic compounds which have high coordination number, eg. NaCl, KCl, CsCl etc 8 (d) CsCl has a bcc lattice. So, dbody = a 3 or dbody = 3 × 0.4123 nm = 0.7141 nm Topic :- STATES OF MATTER Solutions
The sum of the ionic radii of Cs + and Cl ― ions is half this distance ie rCs+ + rCs― = dbody 2 = 0.7141 2 nm = 0.3571 nm Ionic radius of Cs + = 0.3571 ― 0.181 = 0.1761 9 (b) According to ideal gas equation pV = nRT n = number of moles of gas then, pV nRT = 1 Therefore, the compressibility factor Z = pV nRT = 1 For an ideal gas. For real gas Z may be either greater than one or less than one. 10 (a) pV nRT > 1, the gas is less compressible than expected from ideal behaviour and shows positive deviation. 11 (b) PV = w m RT 12 (c) Given, p2 p1 = 2, T2 T1 = 2, V1 = 4 dm3 , V2 = ? From gas equation p1V1 T1 = p2V2 T2 or V1 V2 = p2 p1 × T1/T2 ∴ 4 V2 = 2 × 1 2 = 1 ∴ V2 = 4 dm3 13 (b) A principle used for cooling gas.