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JEE Main 20178 AprilOnline
Exam
JEE Main
Year
2017
Date
8 April
Mode
Online
Subject
Chemistry
Chapter
Solutions

JEE Main 2017 · 8 April · Online · Chemistry · Solutions

5g of Na _ 2 SO _ 4 was dissolved in xg of H _ 2 O . The change in...

5g5g of Na2SO4{Na}_{2}{SO}_{4} was dissolved in xgxg of H2O{H}_{2}O . The change in freezing point was found to be 3.82oC3{.82}^{o}C . If Na2SO4{Na}_{2}{SO}_{4} is 81.581 . 5 % ionised, the value of xx

( Kf{K}_{f} for water =1.86oCkgmol−1= 1{.86}^{o}Ckg{mol}^{- 1} ) is approximately:

(molar mass of S=32gmol−1S = 32g{mol}^{- 1} and that of Na=23gmol−1Na = 23g{mol}^{- 1} )

Options

  1. A25g25g
  2. B65g65g
  3. C15g15g
  4. D45g45g

Correct answer

D. 45g45g

Step-by-step solution

Na2SO4→2Na++SO42−{Na}_{2}{SO}_{4} \to 2{Na}^{+} + {SO}_{4}^{2 -}

i=1+(3−1)0.815=2.63i = 1 + \left(3 - 1\right)0 .815 = 2 .63

3.82=1.86×2.63×5×1000142×x3 .82 = 1 .86 \times 2 .63 \times \dfrac{5 \times 1000}{142 \times x}

∴x=1.86×2.63×5000142×3.82∴ x = \dfrac{1 .86 \times 2 .63 \times 5000}{142 \times 3 .82}

=45gm= 45gm
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