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\(\mathrm{H}_2 \mathrm{~S}\), a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of \(\mathrm{H}_2 \mathrm{~S}\) in water at STP is \(0.195 \mathrm{~m}\), calculate Henry's law constant.
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Solubility of \(\mathrm{H}_2 \mathrm{~S}\) gas \(=0.195 \mathrm{~m}\)
\(=0.195\) mole in \(1 \mathrm{~kg}\) of solvent
\(1 \mathrm{~kg}\) of solvent \(=1000 \mathrm{~g}\)
\(=\frac{1000}{18}=55.55\) moles
\(\therefore x_{\mathrm{H}_2 \mathrm{~S}}=\frac{0.195}{0.195+55.55}\)
\(=\frac{0.195}{55.745}=0.0035\)
Pressure at \(\mathrm{STP}(\mathrm{P})=0.987\) bar
Applying Henry's law,
\(\begin{aligned}
& P_{\mathrm{H}_2 \mathrm{~S}}=K_{\mathrm{H}} \times x_{\mathrm{H}_2 \mathrm{~S}} \\
\Rightarrow \quad K_{\mathrm{H}} &=\frac{P_{\mathrm{H}_2 \mathrm{~S}}}{x_{\mathrm{H}_2 \mathrm{~S}}}=\frac{0.987}{0.0035}=282 \mathrm{bar}
\end{aligned}\)
\(=0.195\) mole in \(1 \mathrm{~kg}\) of solvent
\(1 \mathrm{~kg}\) of solvent \(=1000 \mathrm{~g}\)
\(=\frac{1000}{18}=55.55\) moles
\(\therefore x_{\mathrm{H}_2 \mathrm{~S}}=\frac{0.195}{0.195+55.55}\)
\(=\frac{0.195}{55.745}=0.0035\)
Pressure at \(\mathrm{STP}(\mathrm{P})=0.987\) bar
Applying Henry's law,
\(\begin{aligned}
& P_{\mathrm{H}_2 \mathrm{~S}}=K_{\mathrm{H}} \times x_{\mathrm{H}_2 \mathrm{~S}} \\
\Rightarrow \quad K_{\mathrm{H}} &=\frac{P_{\mathrm{H}_2 \mathrm{~S}}}{x_{\mathrm{H}_2 \mathrm{~S}}}=\frac{0.987}{0.0035}=282 \mathrm{bar}
\end{aligned}\)
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