18 g glucose $(C_{6} H_{12} O_{6})$ is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous
solution is:

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752.4

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759

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7.6

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Solution

The correct option is B 752.4
Moles of glucose ( $C_{6} H_{12} O_{6}) = 0.1$

Moles of water $H_{2} O = \frac{178.2}{18} = 9.9$
From Raoult's law

$\frac{p^{0} - p}{p^{0}} = \frac{0.1}{10}$

$\Rightarrow p^{0} - p = \frac{0.1}{10} \times 760 = 7.6$

$\Rightarrow p = p^{0} - 7.6$

$\Rightarrow p = 760 - 7.6$

$= 752.4 torr$

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Similar questions

(C_{6} H_{12} O_{6})

100^{\circ} C

18 g

(C_{6} H_{12} O_{6})

178.2 g

760

18 g glucose $(C_{6} H_{12} O_{6})$ is added to 178.2 g water. The vapor presure of water (in torr) for this aqueous solution is:

18 g of glucose is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100º C is:

18

C_{6} H_{12} O_{6}

178.2

100^{0} C

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