Question

Henry's law constant for $N_2$ at 310 K is 82.35 kbar. $N_2$ exerts a partial pressure of 0.0840 bar. If $N_2$ gas is bubbled through water at 293 K, then the number of millimoles of $N_2$ that will dissolven in 1 L of water is

• A. $7.16\times {10}^{-2}$
• B. $1.30\times {10}^{-5}$
• C. $1.25\times {10}^{-2}$
• D. $5.55\times {10}^{-2}$

Answer 1

The correct option is D $5.55\times {10}^{-2}$

Henry's law constant is in the unit of pressure, hence we use relation

$p=K_H{\chi }_{N_2}$

${\chi }_{N_2}=\frac{p}{K_H}=\frac{0.840bar}{82.35\times {10}^{3}bar}=1.0\times {10}^{-5}$

$1L{H}_{2}O=1000mL{H}_{2}O=1000g{H}_{2}O(\because d_{H_{2}O}=1gm{L}^{-3})$
$=\frac{1000}{18}=55.5\text{moles}$

Let the number of moles of nitrogen be n

then ${\chi }_{N_2}=\frac{n_{N_2}}{n_{N_2+55.5}}\simeq \frac{n_{N_2}}{55.5}$ (Since, $n_{N_2}$ is very small)

$\therefore =55.5\times 1.0\times {10}^{-5}$

$=5.55\times {10}^{-4}\text{mol}=5.55\times {10}^{-4}\times 1000$
$=5.55\times {10}^{-2}\text{millimole}=0.0555\text{millimole}$
Thus option (d) is correct.