Question

Calculate the vapor pressure of an aqueous solution of 1 molal glucose solution at 100 degrees Celsius.

Answer 1

Step 1: Given data

$\mathrm{Molality}=1\mathrm{m}$;

$\mathrm{Temperature}={100}^{\mathrm{o}}\mathrm{C}$

Step:2 Introducing the Formula:

$\frac{P^o-P}{P^o}=\frac{W_2}{M_2}\times \frac{M_1}{W_1}$

Step: 3 Calculating the mass of solvent using the given molality:

$$\begin{gathered} molality=\frac{W_2}{M_2\times W_1}\times 1000 \\ 1m=\frac{W_2}{M_2\times W_1}\times 1000 \\ \frac{1}{1000}=\frac{W_2}{M_2\times W_1} \\ \frac{W_2}{M_2\times W_1}=0.001 \end{gathered}$$

Step:4 Putting all the values in the above-mentioned equation:

$$\begin{gathered} \frac{760-\mathrm{P}}{760}=\left(\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}\times \frac{1}{{\mathrm{W}}_1}\right)\times {\mathrm{M}}_1 \\ \frac{760-\mathrm{P}}{760}=\left(0.001\right)\times 18=0.018 \\ 760-\mathrm{P}=0.018\times 760=13.68 \\ \mathrm{P}=760-13.78=746.32\mathrm{unit} \end{gathered}$$

Therefore, the vapor pressure of an aqueous solution of 1 molal glucose solution at 100 degrees Celsius is $746.32units.$