Question

A mixture of nitrogen and water vapours is admitted to a flask which contains a solid drying agent immediately after admission, the pressure of the flask is 760 mm. After some hours the pressure reached a steady value of 745 mm.
(a) Calculate the composition, in mol and percent of original mixture.
(b) If the experiment is done at ${20}^{0}C$ and the drying agent increases in weight by 0.15 g. What is the volume of the flask? (the volume occupied by the drying agent may be ignored)?

• A. $3.8$
• B. $2$L
• C. $10.2$L
• D. none

Answer 1

The correct option is C $10.2$L

$\begin{array}{c}\text{Given,}\\ P_{N_2}+P_{H_{2}O}=760mm\\ H_{2}O\text{is dried by drying agent}\\ \text{and lef pressure stands}\\ \text{for}{N}_2\text{.}\\ \text{Thus,}{P}_{N_2}=745mm\\ P_{H_{2}O}=760-745=15mm\end{array}$

percentage of $N_2$ in the

mixture: $\frac{745}{760}\times 100=98.03\%$

percentage $H_{2}O=\frac{15}{760}\times 100$

$=1.97\%$

The mass of $H_{2}O$ in mixture

$=$ increase in mass of drying

agent

$=0.15g$

. . Mole of $H_{2}O\left(x\right)=\frac{0.15}{18}$

Now, use $PV=nRT$ for water

vapours, in a flask of

volume $V$.

$\begin{array}{cc}\frac{15}{760}\times V=\frac{0.15}{18}\times 0.0821\times 293& \\ =& 10.20L[\begin{array}{c}T=20\\ +273\\ =293k\end{array}\\ \therefore V=10.20L& \end{array}$