Question

The molality of $1L$ solution with $x\%H_{2}S{O}_4(w/v)$ is equal to $9.$ The weight of solvent present in the solution is $910g$. The value of $x$ is-

• A. $90$
• B. $80.3$
• C. $100$
• D. $95.5$

Answer 1

Answer: (B)

molality $=9$ which means,

9 moles of solute $\left(H_{2}S{O}_4\right)$ are present per $kg$ of solvent.

So, no. of moles of solute in $1000g$ solvent $=9$

no. of moles of solute in $910g$ solvent $=\frac{9}{1000}\times 910$

$=\frac{819}{100}$

$=8.19$ moles

wt. of 8.19 moles of $H_{2}S{O}_4=98\times 8.19=802.62g$

$\%$ age $(w/v)$ of $H_{2}S{O}_4=\frac{802.62g}{1000ml}\times 100=80.26\%$