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
B
752.4
C
759
D
7.6

Solution

## The correct option is B 752.4The 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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