Solutions Question 371
Question: On mixing 3 g of non - volatile solute in 200 mL of water, its boiling point $ ( 100{}^\circ ) $ becomes $ 100.52{}^\circ C. $ If $ K_{b} $ for water is 0.6 K/m then molecular wt. of solute is:
Options:
A) $ 10.5gmo{l^{-1}} $
B) $ 12.6gmo{l^{-1}} $
C) $ 15.7,gmo{l^{-1}} $
D) $ 17.3,gmo{l^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \Delta T_{b}=K_{b}\frac{w}{M}\times \frac{1000}{W} $
$ 0.52=0.6\times \frac{3}{m}\times \frac{1000}{200}(W=200\times 1) $
$ m=\frac{1.8\times 5}{0.52}=17.3g,mo{l^{-1}} $
