Question

Molarity of a $50mL{H}_{2}S{O}_4$ solution is $10.0M$. If the density of the solution is $1.4g/cc$, calculate its molality

• A. 7.14
• B. 8.00
• C. 10.0
• D. 23.8

Answer 1

Answer: (D)

Given, molarity of $H_{2}S{O}_4$ solution $=10.0M$

Volume $(V)=50mL=.05L$

Density of solution $(d)=1.4g/cc$

Molarity $10m$ means 1 mole of $H_{2}S{O}_4$ present in IL ( $1000mL$ ) of solution. Also, molar mass of $H_{2}S{O}_4\left(M_B\right)=98$

$\because 1000mL$ has $H_{2}S{O}_4=980g(98\times 10)$

$\therefore 50mL$ has $H_{2}S{O}_4=\frac{980\times 50}{1000}=49g$

Thus, mass of solute $\left(w_B\right)=49g$

Calculation for mass of solution and solvent:

$\because d=\frac{\text{ mass of solution }}{\text{ volume ofsolution }}$

$\therefore$ Mass of solution $=d\times V=1.4\times 50=70g$

Thus, mass of solvent $\left(w_A\right)=$ mass of solution-mass of solute. $=70-49$

$w_A=21g$

Calculation for molarity $(m)$ :

$m=\frac{w_B}{M_B}\times \frac{1000}{w_{Akg}}=\frac{49}{98}\times \frac{1000}{21}=23.80m$

Molarity $(m)=23.8m$