Question

The number of moles of solute present in the solutions of $I, II$ and $III$ is respectively
I. $500 \,mL$ of $0.2\, M\, NaOH$
II. $200 \,mL$ of $0.1 \,N \,H _{2} SO _{4}$
II. $6 g$ of urea in $1 \,kg$ 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}=0.1 \,mol $

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

Molarity $\left( H _{2} SO _{4}\right)=\frac{0.1}{2}=0.05\, M$

moles of $H _{2} SO _{4}=\frac{0.1 \times 200}{1000 \times 2}=0.01 \,M $

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

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