A 1.5 m solution of acetic acid (I) is mixed with 3m solution of the acetic acid (II) to prepare 2m solution (m is molality of solution). Select the correct statement(s) : (I) Mass ratio of solvents m

Question

A 1.5 m solution of acetic acid (I) is mixed with 3m solution of the acetic acid (II) to prepare 2m solution (m is molality of solution). Select the correct statement(s) :

(I) Mass ratio of solvents mixed $(\frac{I}{II})$ is $\frac{1}{2}$

(II) Mass ratio of solvents mixed ( $\frac{I}{II}$ ) is $\frac{2}{1}$

(III) Mass ratio of solutions mixed ( $\frac{I}{II}$ ) is $\frac{109}{59}$

(IV) Mass ratio of solutions mixed ( $\frac{I}{II}$ ) is $\frac{59}{109}$

TestHub · Accepted Answer

Let mass of solvent in 1.5 m solution = m 1 ​ Kg and let mass of solvent in 3 m solution = m 2 ​ Kg . Therefore molality of resulting solution = m 1 ​ + m 2 ​ 1.5 m 1 ​ + 3 m 2 ​ ​ = 2 . ∴ m 2 ​ m 1 ​ ​ = 1 2 ​ Therefore required ratio = 59 109 ​

Chemistry - Concentration Terms Question with Solution | TestHub

Options:

Doubts & Discussion