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The vapour pressure of a saturated solution of sparingly soluble salt `("XCl"_(3))` was 17.20mm Hg at `27^(@)"C"`. If the vapour pressure of pure `"H"_(2)"O"` is 17.25 mm Hg at 300K, what is the solubility of sparingly soluble salt `"XCl"_(3)" in mole"//"Litre".`A. `4.04xx10^(-2)`B. `8.08xx10^(-2)`C. `2.032xx10^(-2)`D. `4.04xx10^(-2)` |
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Answer» Correct Answer - A Let solubility of `"XCl"_(3)="S mol lit"^(-1)` For `"XCl"_(3),"n"=4,alpha=1` `therefore"i"=4` As S is very small `therefore` Molality=Molarity `("P"^(0)-"P"_("solution"))/("P"^(0))xx"i"xx("n"_("solute"))/("n"_("solvent")` `(17.25-17.20)/17.25=(4"S")/55.55` `"S"=4.04xx10^(-2)"mol lit"^(-1)` |
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