Question

Asked in: BITSAT-2018

18 g of glucose \left ( C_{6}H_{12}O_{6} \right ) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100^{\circ}C is

A.

759.00 torr

B.

7.60 torr

C.

76.00 torr

D.

752.40 torr.

Answers (1)

The molecular mass of water =2×1+1×16=18g=2×1+1×16=18g

For 178.2g water n_A=9.9

The molecular mass of glucose = 12×1+6×16=180g

For 18g glucose n_B=0.1

X_{B}=\frac{0.1}{0.1+9.9 )}=0.01

X_{A}=0.99

For lowering the vapor pressure,

\begin{aligned} p &=p_{A}^{0} X^{A}=P_{A}^{0}\left(1-X_{B}\right) \\ P &=760(1-0.01) \\ &=760-7.6 \\ &=752.4 \text { torr } \end{aligned}

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