Question

At $0^{\circ} C$ and an $O _{2}$ pressure of $2.00 ,atm$, the aqueous solubility of $O _{2}( g )$, is $60.2 \,mL O _{2}$ per litre. Thus, molarity of $O _{2}$ in a saturated water solution when the $O _{2}$ is under the normal partial pressure of $0.3024 \,atm$, 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 $0^{\circ} C =\frac{p V}{R T}$

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

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

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

Thus, molarity of $O _{2}$ gas at $1 \,atm $

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

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 $

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