Question

The boiling point and freezing point of a solvent A are $$90.0^0 C$$ and $$3.5^0 C$$ respectively. $$K_f$$ and $$K_b$$ values of the solvent are 17.5 and 5.0 K.kg/mol respectively. What is the boiling point of a solution of B in A if the solution freezes at $$2.8^0 C$$?

A
90.00C
B
89.80
C
90.20C
D
90.70C

Solution

The correct option is D $$90.2^0 C$$As we know $$\Delta T_{b} = k_{b}m$$$$\Delta T_{f} = k_{f}m$$Given boiling point = $$90^{0}C$$freezing point = $$3.5^{0}C$$$$k_{b} = 5 \ K/m$$$$k_{f} = 17.5 \ K/m$$Dividing both equations, we get $$\dfrac{T - 90}{3.5-2.8} = \dfrac{5}{17.5}$$After calculating, we get :$$T = 90.2^{0}C$$Chemistry

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