Question

An aqueous solution of glucose is made by dissolving 10g of glucose in 90g water at 303 K. if the V.P.of pure water at 303 K be 32.8mmHg, what would be VP of the solution?

Answer 1

sorry for bad pic...............

Answer 2

To determine the required answer, we would require the following information:

Molar mass of Glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)=180 \mathrm{gm} / \mathrm{mol}$

Molar mass of Water $\left(\mathrm{H}_{2} \mathrm{O}\right)=18 \mathrm{gm} / \mathrm{mol}$

Therefore, the number of moles of sugar in the given quantity  

$\begin{array}{l}{=\frac{\text {Given mass}}{\text {molarmass}}=\frac{10}{180} \mathrm{mole}} \\ {=0.05 \mathrm{mole}}\end{array}$

Moles of Water present in given quantity $=\frac{90}{18}\ \mathrm{mole}$

= 5 mole

Now we need to determine the mole fraction of resulting solution  

= moles of water present / total number of moles present in solution (water + glucose)  

$\begin{aligned} &=\frac{5}{(5+0.5)} \\ &=\frac{5}{5.05} \\ &=0.99 \end{aligned}$

Next we have to determine the vapour pressure of the resultant solution as follows:

Where V.P means Vapour Pressure.

$\begin{array}{l}{\text { V.P of solution }=\text { Mole fraction solvent (water) } \times \text { Pure Vapour Pressure of solvent }} \\ {\text { (water) }}\end{array}$

$\begin{aligned} \mathrm{VP} \text { of solution } &=(0.99 \times 32.8) \mathrm{mm} . \text { of } \mathrm{Hg} \text { . } \\ &=32.47 \mathrm{mm} . \text { of } \mathrm{Hg} \end{aligned}$