Solutions Question 256

Question: The boiling point elevation constant for benzene is $ 2.57{}^\circ C/m. $ The boiling point of benzene is $ 81{}^\circ C $ . Determine the boiling point of solution formed when 10 g of $ C_4H_{12} $ is dissolved in 20 g benzene.

Options:

A) 71.46

B) 7.14

C) 85.76

D) 88.14

Show Answer

Answer:

Correct Answer: D

Solution:

$ \Delta T=k_{b}\cdot m $

$ =k_{b}\cdot \frac{W_{B}}{{M_{W_{B}}}}\times \frac{1000}{W_{A}} $

$ =2.57\times \frac{10}{180}\times \frac{1000}{20} $

$ ={{7.14}^{0}}C $ Now, $ \Delta T=T_{b}-T_b^{0} $
$ \Rightarrow T_{b}=\Delta T_{b}+T_b^{0} $

$ =7.14+81=88.14{}^\circ C $



NCERT Chapter Video Solution

Dual Pane