Question

Molarity of a $50\, mL\, H _{2} SO _{4}$ solution is $10.0 \,M$. If the density of the solution is $1.4 \,g / cc$, calculate its molality

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

Answer 1

Answer: (D)

Given, molarity of $H _{2} SO _{4}$ solution $=10.0\, M$

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

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

Molarity $10\, m$ means 1 mole of $H _{2} SO _{4}$ present in IL ( $1000 mL$ ) of solution. Also, molar mass of $H _{2} SO _{4}\left(M_{B}\right)=98$

$\because \,1000 \,mL$ has $H _{2} SO _{4}=980 \,g (98 \times 10)$

$\therefore \,50 \,mL$ has $ H _{2} SO _{4}=\frac{980 \times 50}{1000}=49\, g$

Thus, mass of solute $\left(w_{B}\right)=49 \,g$

Calculation for mass of solution and solvent:

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

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

Thus, mass of solvent $\left(w_{A}\right)=$ mass of solution-mass of solute. $=70-49$

$w_{A}=21 g$

Calculation for molarity $(m)$ :

$m=\frac{w_{B}}{M_{B}} \times \frac{1000}{w_{A k g}}=\frac{49}{98} \times \frac{1000}{21}=23.80\, m$

Molarity $(m)=23.8 \,m$