A solution contains 20 g glucose C6H12O6 and 34.2 g of sucrose C12H22O11 in 108 g water. If the vapour pressure of pure water is 35 mmHg, the decrease in vapour pressure of the solvent will be

A
$0.85 mmHg$
B
$0.98 mmHg$
C
$1.19 mmHg$
D
$2.25 mmHg$

Solution:

Amount of glucose, $n_{2} = \frac{m_{2}}{M_{2}} = \frac{20 g}{180 g {mol}^{- 1}} = \frac{1}{9} mol$ ; Amount of sucrose, $n_{3} = \frac{m_{3}}{M_{3}} = \frac{34.2 g}{342 g {mol}^{- 1}} = \frac{1}{10} mol$

Amount of water, $n_{1} = \frac{m_{1}}{M_{1}} = \frac{108 g}{18 g {mol}^{- 1}} = 6 mol$

Mole fraction of solutes in solution,

$x = \frac{n_{2} + n_{3}}{n_{1} + n_{2} + n_{3}} = \frac{[(1 / 9) + (1 / 10)] mol}{6 mol + ⌊ (1 / 9) + (1 / 10) mol]} = \frac{0.211 mol}{6.211 mol} = 0.034$

Vapour pressure of solution. $p_{1} = x_{2} p_{1}^{*} = (0.034) (35 mmHg) = 1.19 mmHg$

A solution contains 20 g glucose $(C_{6} H_{12} O_{6})$ and 34.2 g of sucrose $(C_{12} H_{22} O_{11})$ in 108 g water. If the vapour pressure of pure water is 35 mmHg, the decrease in vapour pressure of the solvent will be

Solution:

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