Question

if mole fraction of solute is an aqueous solution is changed from 1/2 to 1/4 by adding water, then the ratio of initial and final molality of solute in solution is?
a 1:1
b 2:1
c 3:1
d 4:1​

Answer 1

Answer:

b) 2:1

Explanation:

initial / final

(1/2)/(1/4)

4/2

2:1

Answer 2

Given:

Change in mole fraction of solute from 1/2 to 1/4

To Find: Ratio of initial and final molality of solute in solution

Solution:

Since the mole fraction of solute initially is 1/2,

Therefore, moles of the solute $\[ = \frac{1}{2} \times \]$ total number of moles in the solution

$\[{n_{solute}}_1 = \frac{1}{2} \times ({n_{solute1}} + {n_{water}})\]$

$\[{n_{solute}}_1 - \frac{1}{2} \times {n_{solute1}} = \frac{1}{2}{n_{water}}\]$   ... (i)

Similarly, when the mole fraction of solute becomes 1/4

$\[{n_{solute}}_2 = \frac{1}{4} \times ({n_{solute2}} + {n_{water}})\]$

$\[{n_{solute}}_2 - \frac{1}{4} \times {n_{solute2}} = \frac{1}{4}{n_{water}}\]$   ... (ii)

Therefore, the ratio of molality can be given as:

$\[\frac{{{M_1}}}{{{M_2}}} = \frac{{{n_{solute}}_1}}{{{n_{solute}}_2}}\]$

$\[\frac{{{M_1}}}{{{M_2}}} = \frac{{0.5{n_{water}}/0.5}}{{0.25{n_{water}}/0.75}}\]$

$\[\frac{{{M_1}}}{{{M_2}}} = \frac{{0.75}}{{0.25}}\]$

$\[\frac{{{M_1}}}{{{M_2}}} = \frac{3}{1}\]$

Thus,

$\[{M_1}:{M_2} = 3:1\]$

Hence, the initial and final molality of solute in the solution is 3:1. Thus, the correct option is c.