Question

What volume of $5M$ $N{a}_{2}S{O}_4$ must be added to $25mL$ of $1M$ $BaC{l}_2$ to produce $10g$ of $BaS{O}_4$?

• A. $8.57mL$
• B. $7.2mL$
• C. $10mL$
• D. $12mL$

Answer 1

Answer: (A)

$8.57mL$

$N{a}_{2}S{O}_4+BaC{l}_2\to 2NaCl+BaS{O}_4$

Molar mass of $N{a}_{2}S{O}_4$=142 g/mol,$BaS{O}_4$=233.3 g/mol, $BaC{l}_2$=208.3 g/mol

number of moles=volume$\times$molarity=mass/molar mass

1 mol of $N{a}_{2}S{O}_4$ reacts with 1 mol of $BaC{l}_2$ to give 1 mol of $BaS{O}_4$

No. of moles of the sulfate ion = no. of moles of barium ion in $BaS{O}_4$

Number of moles of barium ion in 10 g barium sulphate=number of moles of $BaS{O}_4=\frac{10}{233.40}=0.0428$ moles

$\therefore$ no. of moles of sulphate ions= 0.0428 mole

Also, number of moles of sulphate ion =number of moles of $N{a}_{2}S{O}_4$ =$M\times V=5\times V=0.0428$

$V=8.57\times {10}^{-3}L=8.57mL$