`H_(2)S`gas is used in qualitative analysis of inorganic cations. Its solubility in water at STP is `0.195 mol Kg^(-1)`.Thus, Henry's law constant (in atm mola`l^(-1)`)for`H_(2)S` is

Options

A. 386 bar
B. 195 bar
C. 478 bar
D. 282 bar

Answer

4

Explanation

Applying Henry's law, we have
`P_(H_(2)O)= K_(H)chi_(H_(2)O)`
or `K_(H)=P_(H_(2)O)/chi_(H_(2)O)`
Thus, we need to find `chi_(H_(2)O)` from its molality. A `0.195 m H_(2)S` solution contains `0.195` mole of `H_(2)S` in `1.00 kg` of water. After converting `1.00 kg H_(2)O` to moles, we can calculate the mole fraction using its definition.
The moles of `H_(2)O` in `1.00 kg` of water are
`n_(H_(2)O)=(mass_(H_(2)O))/(molar mass_(H_(2)O))`
`=(1000 g)/(18 g mol^(-1))`
`=55.6 mol`
Hence, `chi_(H_(2)S)=n_(H_(2)S)/(n_(H_(2)S)+n_(H_(2)S))=(0.195 mol)/((0.195+ 55.6) mol)`
`=(0.195 mol)/(55.8 mol)=0.0035`
Since pressure at STP is 0.987 bar, we
have, `K_(H)=P_(H_(2)S)/chi_(H_(2)S)=(0.987 "bar")/0.0035`
282 bar