Question

The number of moles of solute present in the solutions of $I,II$ and $III$ is respectively
I. $500mL$ of $0.2MNaOH$
II. $200mL$ of $0.1N{H}_{2}S{O}_4$
II. $6g$ of urea in $1kg$ of water

• A. 0.1, 0.01, 0.1
• B. 0.1, 0.02, 0.1
• C. 0.2, 0.01, 0.1
• D. 0.1, 0.01, 0.2

Answer 1

Answer: (A)

I. Moles of solute $NaOH$ = molarity $\times$ volume of solution (in $L$ )

$=\frac{0.2\times 500}{1000}=0.1mol$

II. Normality = n -factor $\times$ molarity

Molarity $\left(H_{2}S{O}_4\right)=\frac{0.1}{2}=0.05M$

moles of $H_{2}S{O}_4=\frac{0.1\times 200}{1000\times 2}=0.01M$

III. Moles $=\frac{\text{ Weight of urea }}{\text{ Molecular weight of urea }}$

$=\frac{6}{60}=0.1m$