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

Option 1)
759.00 torr

Option 2)
7.60 torr

Option 3)
76.00 torr

Option 4)
752.40 torr.

Answers (1)

P perimeter

As we learnt in

Expression of relative lowering of vapour pressure -

$\frac{\Delta P}{ P^{0}}= x_{solute}$

$x_{solute}= \frac{ n_{solute}}{ n_{solute}+n_{solvent}}$

- wherein

$\Delta P \: is \: lowering \: o\! f \: v.p.$

$P^{0}\rightarrow \: vapour\: pressure\: of \: pure\: solvent$

$x_{solute}\rightarrow \: mole\: fraction \: of \:non\: volatile\: solute$

$\frac{p_{A}^{0}-p_{s}}{p_{A}^{0}}=\chi_{B}$

Mole fraction of glucose $(\chi_{B})=\frac{18/180}{\frac{18}{180}+\frac{178.2}{18}}=\frac{1}{100}$

$p_{A}^{0}=$ Vapour pressure of pure water at 100 c^o = 760 torr

$\frac{760-p_{s}}{760}=\frac{1}{100}$

p_s = Vapour pressure of solution = 752.4 torr.

Correct option is 4.

Option 1)

759.00 torr

This is an incorrect option.

Option 2)

7.60 torr

This is an incorrect option.

Option 3)

76.00 torr

This is an incorrect option.

Option 4)

752.40 torr.

This is the correct option.

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18 g of glucose is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at is Option 1) 759.00 torr Option 2) 7.60 torr Option 3) 76.00 torr Option 4) 752.40 torr.

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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

Option 1)
759.00 torr

Option 2)
7.60 torr

Option 3)
76.00 torr

Option 4)
752.40 torr.