Question

Density of $2.03M$ aqueous solution of acetic acid is $1.017gm{L}^{-1}$ molecular mass of acetic is $60$. Calculate the molality of the solution.

• A. $2.27$
• B. $1.27$
• C. $3.27$
• D. $4.27$

Answer 1

The correct option is A $2.27$

Given,

Molarity of aqueous solution of acetic acid=$2.03M$

Density=$1.017gm{L}^{-1}$

Also, Molar mass of acetic acid=$60gmo{l}^{-1}$

We know,

Molality=$\frac{No.ofmolesofsolute}{massofsolventinKg}⟶1$

Molarity=No. of moles in $1L$ solution

$\therefore$ No. of moles of acetic acid in $1000ml$ solution=$2.03$

Now, Density=$\frac{mass}{volume}\Rightarrow mass=density\times volume$

$\Rightarrow$ Mass of solutuion=$1.017\times 1000=1017g⟶2$

Now, Mass of acetic acid in the solution=No. of moles of $C{H}_{3}COOH\times$ Molar mass of $C{H}_{3}COOH$

$=2.03\times 60=121.8g⟶3$

$Rightarrow$ Mass of solvent=Mass of solution-Mass of $C{H}_{3}COOH$

$=1017g-121.8g$

$=895.2g$ $[\because 2$ & $3]$

From 1, Molality of solution=$\frac{2.03}{895.2\times {10}^{-3}}$

$=2.27m$