Question

Suppose a glucose solution was prepared by dissolving 5.00 g of glucose in 100 g of water. At 363 K (90 degrees celcius). The experimentally determined vapor pressure of the glucose solution was found to be 69727.60 Pa. The vapor pressure of water at 363 K is 70100.90 Pa. Calculate the molar mass of glucose.

Answer 1

number of moles of glucose = $\dfrac{\textsf{mass}}{\textsf{molar mass}}$ = $\dfrac{5}{180}$ = 0.028 moles

number of moles of water = $\dfrac{100}{18}$ = 5.556 moles

Total number of moles of water = 5.556 + 0.028 = 5.584

molar fraction of solvent = $\dfrac{5.556}{5.584}$ = 0.995

pressure fraction of solvent = $\dfrac{69727.60}{70100.90}$ = 0.995.

This means the molar mass of glucose is truly 180 g/mol.