Answer : The molar mass of the solute will be 87.90 g/mol.
Explanation : We know the formula for elevation in boiling point, which is
Δt = i[tex] K_{b} [/tex]m
given that, Δt = 0.357, [tex] K_{b} [/tex] = 5.02 and mass of [tex]CCl _{4} [/tex] = 40,
on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X[tex] 10^{-3} [/tex].
Now, 0.250/ 2.844 X[tex] 10^{-3} [/tex] = 87.90 g/mol. which is the weight of unknown component.
