Question

17) Molality of 30% w/w aqueous solution of NaOH is -
(A) 10.71 m
30
30x 10
(B) 7.5 m
(C) 8.32 m
(D) 9.17 m​

Answer 1

Answer:

C

8.32

as it is a whole 30 out of 100

Answer 2

Molality of 30% w/w aqueous solution of NaOH is 6.25 molal .

Given :

w/w aqueous solution of NaOH 30 % .

Let us take 100 gram of solution .

So , mass of NaOH is 20 gram and mass of solvent is 80 gram .

We know , molecular mass of NaOH is 40 gram .

Therefore , number of moles of NaOH is $\dfrac{20}{40}=0.5\ moles\ .$

Now , molality ,

$m=\dfrac{no\ of\ moles\times 1000}{mass\ of\ solvent(in\ gram)}\\\\m=\dfrac{0.5\times 1000}{80}\\\\m=6.25\ molal$

Hence , this is the required solution.

Learn More :

Volumetric Analysis

https://brainly.in/question/12135610