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(a) When 2.56 g of sulphur was dissolved in 100 g of $CS_{2},$ the freezing point lowered by 0.383 K. Calculate the formula of sulphur $(S_{X})$

( $K_{f}$ for $CS_{2}=3.83\; K\; kg\; mol^{-1}$ Atomic mass of sulphur $=32\; g\; mol^{-1}$ )

(b) Blood cells are isotonic with $0.9\; ^{o}/_{o}$ sodium chloride solution. What happens if we place blood cells in a solution containing ?

(i) $1.2\; ^{o}/_{o}$ sodium chloride solution

(ii) $0.4\; ^{o}/_{o}$ sodium chloride solution

Answers (1)

(a) $K_{f}(CS_{2})=3.83\; K\; kg\; mol^{-1}$

Atomic mass of sulphur $=32\; g\; mol^{-1}$

Mass of sulphur $=2.56\; g$

Mass of $CS_{2}=100 \; g$

$\Delta T_{f}=0.383 \; K$

Formula of sulphur $(S_{X}), \; X=?$

$\Delta T_{f}=l\times K_{f}\times m$

$0.383=i\times 3.83\times \frac{2.56\times 1000}{32\times 100}$

$i=\frac{0.383\times 32\times 100}{3.83\times 2.56\times 1000}$

$i=\frac{1}{8}$

$\therefore \; X=8$

Formula of sulphur $=S_{8}$

(b) Blood cells are isotonic with $0.9\; ^{o}/_{o}\; NaCl$ solution . If we place blood cells in -

(i) $1.2\; ^{o}/_{o}\; NaCl$ solution blood cells will shrink.

(ii) $0.4\; ^{o}/_{o}\; NaCl$ solution

blood cells will swell.

Posted by

Sumit Saini

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