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Chemistry

  • Date Submitted: 08/26/2013 11:36 AM
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UNIT-2   :     SOLUTIONS
INTEXT QUESTIONS

2.1: Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
A. Mass percentage of C6H6 

Mass percentage of CCl4

Alternatively,   Mass percentage of CCl4 = (100 − 15.28)%   = 84.72%
2.2:   Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
A.   Let the total mass of the solution be 100 g and the mass of benzene be 30 g. ∴Mass of carbon tetrachloride = (100 − 30)g
= 70 g
Molar mass of benzene (C6H6) =   78 g mol−1
∴Number of moles of    = 0.3846 mol
Molar mass of carbon tetrachloride (CCl4) =   154 g mol−1
∴Number of moles of CCl4  = 0.4545 mol
Thus, the mole fraction of C6H6 is given as:

      = 0.458
2.3:   Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4diluted to 500 mL.
A.   Molarity is given by:

(a) Molar mass of Co (NO3)2.6H2O =   291 g mol−1
∴Moles of Co (NO3)2.6H2O   = 0.103 mol
Therefore, molarity      = 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol
∴Number of moles present in 30 mL of 0.5 M H2SO4    = 0.015 mol
Therefore, molarity   = 0.03 M
2.4:   Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
A.   Molar mass of urea (NH2CONH2) = 60 g mol−1
0.25 molal aqueous solution of urea means:
1000 g of water contains 0.25 mol = (0.25 × 60)g of urea   = 15 g of urea
That is,   (1000 + 15) g of solution contains 15 g of urea
Therefore, 2.5 kg (2500 g) of solution contains    = 36.95 g
= 37 g of urea (approximately)
Hence, mass of urea required = 37 g
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
2.5: Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
A.   (a) Molar mass of KI = 39 + 127 = 166 g...

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