Saved Bookmarks
| 1. |
H_(2)S a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H_(2)S in water at STP is 0.195 m. Calculate Henry's law constant. |
|
Answer» Solution :Solubility of `H_(2)S` gas `=0.195 m` `=0.195` mole in 1 kg of the SOLVENT (water) 1 kg of the solvent (water) `= 1000 gh = (1000g)/(18gmol^(-1)) = 55.55` moles Mole fraction of `H_(2)S` gas in the solution `(X) = (0.195)/( 0.195 + 55.55) = (0.195)/(55.745) = 0.0035` Pressure at STP `=0.987` bar Applying Henry.s law `P_(H_(2)S) = K_(H) xx x _(H_(2)S)` ` K _(H) = (P _(H_(2)S))/(x_(H_(2)S))= (0. 987 b ar)/(0.0035) = 282` bar |
|