Question

At $0^{\circ }C$ and an $O_2$ pressure of $2.00,atm$, the aqueous solubility of $O_2(g)$, 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. $1.62\times 10^{3}M$

Answer 1

Answer: (D)

Moles of $O_2$ at $0^{\circ }C=\frac{pV}{RT}$

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

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

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

Thus, molarity of $O_2$ gas at $1atm$

$=5.37\times 10^{-3}M$

By Henry's law, Concentration $=K_{H}p$

$C_1=K_H{p}_1$

$C_2=K_{H{p}_2}$

$\therefore \frac{C_2}{C_1}=\frac{p_2}{p_1}$

$\therefore C_2=C_1\frac{p_2}{p_1}$

$=\left(5.37\times 10^{-3}\times \frac{0.3024}{1}\right)M$

$=1.62\times 10^{3}M$