Test Index

Solutions

Section: Chemistry

© examsnet.com

Question : 15 of 23

Marks: +1, -0

The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is2.0 g

\mathrm{cm}^{-3}

.The ratio of the molecular weights ofthe solute and solvent,

\left(\frac{\text{MW}_{\text{solute}}}{\text{Mw}_{\text{solvent}}}\right)

[JEE Adv 2016 P1]

Your Answer:

Solution:

1 mole solution has 0.1 mole solute and 0.9 mole solvent

Let

M_1=

Molar mass solute

M_2=

Molar mass solvent

Molality, m

=\frac{0.1}{0.9M_2}\times 1000

....(1)

Molarity M

=\frac{0.1}{0.1M_1+0.9M_2}\times 2\times 1000

........(2)

∵ m =M

\Rightarrow \frac{0.1\times 1000}{0.9M_2} = \frac{200}{0.1M_1+0.9M_2}

\Rightarrow \frac{M_1}{M_2} = 9

Alternate solution

∵ M = m

⇒ volume of solution = mass of solvent

⇒

\frac{W_{\text{solute}} + W_{\text{Solvent}}}{2} = W_{\text{solvent}}

W_{\text{solute}} = W_{\text{solvent}}

0.1\times M_{\text{solute}} = 0.9\times M_{\text{solvent}}

\frac{M_{\text{solute}}}{M_{\text{solvent}}} = 9

© examsnet.com

Go to Question:

Prev Question

Next Question