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Question

18 g glucose $$(C_6H_{12}O_6)$$ is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is: 


A
76
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B
752.4
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C
759
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D
7.6
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Solution

The correct option is B 752.4
The molar masses of glucose and water are 180 g/mol and 18 g/mol respectively.

Number of moles of glucose $$\displaystyle = \dfrac {18}{180} = 0.1 $$

Number of moles of water $$\displaystyle = \dfrac {178.2}{18} = 9.9$$

Mole fraction of glucose $$\displaystyle = \dfrac {0.1}{0.1 + 9.9} = 0.01$$

Relative lowering in vapour pressure is equal to the mole fraction of glucose.

$$\displaystyle \dfrac {760 - P}{760} = 0.01$$

$$\displaystyle 760 - P = 7.6$$

$$\displaystyle P = 760-7.6 = 752.4$$ torr.

Hence, the correct option is $$B$$

Chemistry

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