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Question: What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm H...

What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg?

A

4.69 g

B

3.59 g

C

2.59 g

D

3.69 g

Answer

3.69 g

Explanation

Solution

Raoult's Law states that the relative lowering of vapour pressure is equal to the mole fraction of the solute. For a dilute solution, this is approximated as the ratio of moles of solute to moles of solvent. Using the given values: Po−PPo=Xsolute≈nsolutensolvent\frac{P^o - P}{P^o} = X_{solute} \approx \frac{n_{solute}}{n_{solvent}} 0.20 mm Hg54.2 mm Hg=x/180 g mol−1100 g/18 g mol−1\frac{0.20 \text{ mm Hg}}{54.2 \text{ mm Hg}} = \frac{x/180 \text{ g mol}^{-1}}{100 \text{ g} / 18 \text{ g mol}^{-1}} Solving for xx (weight of glucose): x=0.20×100054.2≈3.69 gx = \frac{0.20 \times 1000}{54.2} \approx 3.69 \text{ g}