Question
The Henry's law constant for the solubility of $${N_2}$$ gas in water at $$298\,K$$ is $$1.0 \times {10^5}\,atm.$$ The mole fraction of $${N_2}$$ in air is $$0.8.$$ The number of moles of $${N_2}$$ from air dissolved in $$10\,moles$$ of water at $$298\,K$$ and $$5\,atm$$ pressure is
A.
$$4.0 \times {10^{ - 4}}$$
B.
$$4.0 \times {10^{ - 5}}$$
C.
$$5.0 \times {10^{ - 4}}$$
D.
$$4.0 \times {10^{ - 6}}$$
Answer :
$$4.0 \times {10^{ - 4}}$$
Solution :
According to Henry's law,
$${x_{{N_2}}} \times {K_H} = {p_{{N_2}}}$$ ( $${p_{{N_2}}} = $$ Partial pressure of $${N_2}$$ )
Given, total pressure $$ = 5\,atm,$$ mole fraction of $${N_2} = 0.8$$
∴ Partial pressure of $${N_2} = 0.8 \times 5 = 4$$
$$ \Rightarrow {x_{{N_2}}} \times 1 \times {10^5} = 4 \Rightarrow {x_{{N_2}}} = 4 \times {10^{ - 5}}$$
no. of moles of $${H_2}O,{n_{{H_2}O}} = 10$$
no. of moles of $${N_2},{n_{{N_2}}} = ?$$
$$\eqalign{
& \frac{{{n_{{N_2}}}}}{{{n_{{N_2}}} + {n_{{H_2}O}}}} = {x_{{N_2}}} = 4 \times {10^{ - 5}} \cr
& \Rightarrow \frac{{{n_{{N_2}}}}}{{10 + {n_{{N_2}}}}} = 4 \times {10^{ - 5}} \cr
& \Rightarrow {n_{{N_2}}} = 4 \times {10^{ - 4}}\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,{n_{{N_2}}} < < < 10} \right] \cr} $$