Question

Find the molality of the solution having $93$% $H_2{SO}_4$ ($w/v$) having density $1.84gm{L}^{-1}$ is:

• A. $1.043m$
• B. $0.143m$
• C. $10.43m$
• D. $0.014m$

Answer 1

Answer: (C)

$10.43m$

93 % $H_{2}S{O}_4$ (w/v) means $93gm$ of $H_2{S}^{}{O}_4$ is present in $100mL$ of solution.

Given,

Mass of solute = 93 $gm$

Volume of solution = 100 $mL$

Density of solution = 1.84 $gm{L}^{-1}$

$\therefore Massofsolution=density\times volumeofsolution$

$=1.84\times 100$

$=184gm$

Mass of solution = 184 $gm$

Mass of solvent = 184 - 93 = 91 $gm$

$Molality=\frac{Molesofsolute}{weightofsolvent\left(kg\right)}$

$=\frac{\frac{93}{98}}{\frac{91}{1000}}$

$=\frac{93\times 1000}{98\times 91}$

$=10.43m$

Hence, option $\left(C\right)$ is correct.