Question

The amount of anhydrous $N{a}_{2}C{O}_3$ present in 250 mL of 0.25 M solution is:

• A. $6.625g$
• B. $66.25g$
• C. $12.68g$
• D. $24.35g$

Answer 1

The correct option is A $6.625g$

We know,
Molarity $M=\frac{\text{moles of solute}}{\text{volume of solution (mL)}}\times 1000$
So, moles of solute = Molarity M $\times$ volume of solution
Moles of solute = $0.25\times \frac{250}{1000}$ = 0.0625 g

molar mass of $N{a}_{2}C{O}_3$ = 106 $g/mol$
mass of solute = $0.0625\times 106$ = $6.625g$