Question

in an experiment 18.04 gram of mannitol dissolved in 100g of water.The vapour pressure of water lowered by 0.309 mmHg to 17.535 mmHg calculate molar mass of manitol

Answer 1

Answer: The molar mass of mannitol is 185.78 g/mol.

Explanation:

pressure of the solution= $p_s=17.535 mm Hg$

Pressure of the pure water =$p_o-p_s=0.309 mmHg$

$p_o=0.306 mmHg+17.535 mmHg=17.841 mmhg$

$\frac{p_s}{p_o}=\chi_w=\frac{17.535 mmHg}{17.841 mmHg}=0.9828$

$\chi_w=\frac{n_w}{n_w+n_m}$

$0.9828=\frac{\frac{100 g}{18 g/mol}}{\frac{100 g}{18 g/mol}+n_m}$

Number of moles of mannitol is the solution = $n_m=0.0971 mol$

$n_m=0.0971 mol=\frac{\text{Mass of the mannitol}}{\text{Molar mass of the mannitol}(M)}=\frac{18.04 g}{M}$

The Molar mass of the mannitol =$\frac{18.04 g}{0.0971 mol}=185.78 g/mol$

Answer 2

Answer:

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