Question

Calculate the boiling point of a solution containing 0.45g of camphor (mol. wt. 152) dissolved in 35.4g of acetone (b.p. 56.3°C); K_b per 100 gm of acetone is 17.2°C.

(A) 56.446°C 

(B) 52.401°C 

(C) 56.146°C 

(D) 50.464°C

Answer 1

Correct Option (A) 56.446°C 

Explanation:

Here it is given that, w = 0.45 g , W = 35.4 , m = 152 , Kb = 17.2 per 100gm 

Now we know that ΔT_b = 100 x K_bxW/mX W

(Note that this is expression when Kb is given per 100g of the solvent)

Substituting the values in the above expression.

 ΔT_{b = 100 x 17.2 x 0.45/152 x 35.4 = 0.146}°C

Now we know that

B.P. of solution (T) - B.P. of solvent (T₀) = ΔT

:.  B.P. of solution (T) = B.P. of solvent (T₀) +  ΔT

Hence B.P. of solution = 56.3 + 0.146 = 56.4460C