JEE MAIN - Chemistry (2022 - 29th July Morning Shift - No. 19)
If $$\mathrm{O}_{2}$$ gas is bubbled through water at $$303 \mathrm{~K}$$, the number of millimoles of $$\mathrm{O}_{2}$$ gas that dissolve in 1 litre of water is __________. (Nearest Integer)
(Given : Henry's Law constant for $$\mathrm{O}_{2}$$ at $$303 \mathrm{~K}$$ is $$46.82 \,\mathrm{k}$$ bar and partial pressure of $$\mathrm{O}_{2}=0.920$$ bar)
(Assume solubility of $$\mathrm{O}_{2}$$ in water is too small, nearly negligible)
Answer
1
Explanation
$$\mathrm{P}=\mathrm{K}_{\mathrm{H}} \times \mathrm{X}$$
$0.920 \mathrm{bar}=46.82 \times 10^{3}\, \mathrm{bar} \times \frac{\mathrm{mol} \,\mathrm{of} \,\mathrm{O}_{2}}{\mathrm{~mol} \text { of } \mathrm{H}_{2} \mathrm{O}}$
$$0.920=46.82 \times 10^{3} \times \frac{\text { mol of O}_{2}}{1000 / 18}$$
$$0.920=46.82 \times n_{O_{2}}$$
$$\mathrm{P}=\frac{0.920}{46.82 \times 18}=n_{O_{2}}$$
$$\Rightarrow 1.09 \times 10^{-3} \,n_{O_{2}}$$
$$\Rightarrow \mathrm{m}\, \mathrm{mol}\, \mathrm{of}\, \mathrm{O}_{2}=1$$
$0.920 \mathrm{bar}=46.82 \times 10^{3}\, \mathrm{bar} \times \frac{\mathrm{mol} \,\mathrm{of} \,\mathrm{O}_{2}}{\mathrm{~mol} \text { of } \mathrm{H}_{2} \mathrm{O}}$
$$0.920=46.82 \times 10^{3} \times \frac{\text { mol of O}_{2}}{1000 / 18}$$
$$0.920=46.82 \times n_{O_{2}}$$
$$\mathrm{P}=\frac{0.920}{46.82 \times 18}=n_{O_{2}}$$
$$\Rightarrow 1.09 \times 10^{-3} \,n_{O_{2}}$$
$$\Rightarrow \mathrm{m}\, \mathrm{mol}\, \mathrm{of}\, \mathrm{O}_{2}=1$$
Comments (0)
