CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
89.80
loader
C
90.20C
loader
D
90.70C
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image