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A very small amount of a non-volatile so...

A very small amount of a non-volatile solute (that does not dissociate) is dissolved in `56.8 cm^(3)` of benzene (density `0.889 g cm^(3))`. At room temperature, vapour pressure of this solution is `98.88 mm Hg` while that of benzene is `100 mm Hg` . Find the molality of this solution. If the freezing temperature of this solution is `0.73` degree lower than that of benzene, what is the value of molal the freezing point depression constant of benzene?

Text Solution

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The correct Answer is:
5
Wt. of benzene `= 56.8 xx 0.889 = 50.49 g`
`:.` Moles of benzene `= (50.49)/(78) = 0.647` moles
Now, `(P^(@)-P_(S))/(P_(S)) = (n)/(N)`
`rArr (100-98.88)/(98.88) = (n)/(0.647)`
`:. n = 7.33 xx 10^(-3)"moles"`
Molallity of solution `= ("No.of moles of solute")/("wt. of solvent (in g)") xx 1000`
`= (7.33 xx 10^(-3))/(50.49) xx 1000 = 0.1452 m`
`because Delta T_(f) = K'_(f) xx m`
`:. 0.73 = K'_(f) xx 0.1452`
`:. K'_(f) ~~ 5 K mol kg^(-1)`
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