Question

At $0^{\circ }C$ and $2.00atm$ partial pressure of oxygen, the aqueous solubility of $O_2\left(g\right)$ is $60.2mL$ $O_2$ per litre. Thus, molarity of $O_2$ in a saturated water solution when the $O_2$ is under the normal partial pressure of $0.3024atm$ is

• A. $4.57\times {10}^{-4}M$
• B. $2.18\times {10}^{-3}M$
• C. $4.57\times {10}^{-2}M$
• D. $8.12\times {10}^{-4}M$

Answer 1

The correct option is D $8.12\times {10}^{-4}M$

Number of mol of $O_2$ at $0^{\circ }C$

$=\frac{pV}{RT}=\frac{2.0atm\times \left(\frac{60.2}{1000}\right)L}{0.0821Latmmo{l}^{-1}{K}^{-1}\times 273K}$

$=\frac{2\times 0.0602}{0.0821\times 273}=\frac{0.1204}{22.4133}$

$=5.37\times {10}^{-3}mol\text{in}1L{H}_{2}O$

Thus, molarity of $O_2$ gas at $2atm$ partial pressure of $O_2\left(g\right)$ = $5.37\times {10}^{-3}M$

By Henry's law,

$p$ = $K_{H}x$
Here we have $x$ = molarity of $O_2\left(g\right)$

$p_1=K_H{x}_1$

$p_2=K_H{x}_2$

$\therefore \frac{p_2}{p_1}=\frac{x_2}{x_1}$

$\therefore x_2=x_1\frac{p_2}{p_1}=(5.37\times {10}^{-3}\times \frac{0.3024}{2})$

$x_2=8.12\times {10}^{-4}M$
Hence, (d) is correct.