Question

The molality of the $3\,M$ solution of methanol if the density of the solution is $0.9\, g\, cm^{-3}$ is

• A. 3.73
• B. 3
• C. 3.33
• D. 3.1

Answer 1

Answer: (A)

3.73

Answer 2

$\because$ Molarity $(C)=\frac{W_{B}(\%) \times d \times 10}{M_{B}}$

and

Molality $(m)=\frac{W_{B}(\%) \times 1000}{\left[100-W_{B}(\%)\right] \times M_{B}}$

where, $W_{B}=$ mass of solute

$M_{B} =$ moler mass of solute.

$W_{B} =$ to find

$M_{B} =$mass of $CH _{3} OH$
$=32(12+3+16+1)$
$\therefore W_{B}(\%)=\frac{C \times M_{B}}{d \times 10}$
$=\frac{3 \times 32}{0.9 \times 10}=10.66 \,g (\%)$

and $m=\frac{W_{B}(\%) \times 1000}{[100-10.66] \times 32}=\frac{10.66 \times 1000}{89.34 \times 32}$
$m=3.728 \approx 3.73 \,m$