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Calculatethe boiling point of a one molaraqueous solution (density 1.04g"mL "^(1))ofpotassium chlorideK_(b)for water = 0.52K "mol "^(-1). Atomicmasses : K = 39, Cl = 35.5 ) |
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Answer» Solution :Concentrationof solution = 1M ,Density of solution = 1.04 g `" mL"^(-1)` LET us FIRST calculate the motality of the solution. mount of solute (KCI) = 1mol = 74.5 g , Volumeof solution= 1:L = 1000 mL Mass of the solution ` = 1000 xx 1.04 g = 1040 g` `:. ` Mass of solvent=` 1040 - 74.5` g = 965 . 5 g = 0.9655 kg `"Molality of the solution"= ("No. of moles of the solute ")/("Mass of the solvent in kg")= ("1mol")/(0.9655 "kg") = 1.0357 "molkg"^(-1) = 1.0357 m` KCl DISSOCIATES as : ` KCl to K^(+)Cl^(-)` ` :. ` Numberof particles after dissociation=2`:. `van't Hoff factor , i = 2 Now` Delta T _(b) = i xx K_(b) xx m = 2 xx 0.52 xx 1.0357 = 1.078^(@) C` `:. ` BOILING point of the solution = ` 100 + 1.078 = 101.078 ^(@)C` |
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