Question

What will be the weight of $N{a}_{2}C{O}_3$ of 90% purity required to prepare to neutralize 100 mL of 0.2 N $H_{2}S{O}_4$ ?

• A. $0.66g$
• B. $1.17g$
• C. $3.82g$
• D. $10.69g$

Answer 1

The correct option is B $1.17g$

Meq. of $N{a}_{2}C{O}_3=$ Meq. of $H_{2}S{O}_4$
Meq. of $H_{2}S{O}_4=100\times 0.2$
$\therefore \frac{W_{N{a}_{2}C{O}_3}}{E_{N{a}_{2}C{O}_3}}\times 1000=100\times 0.2$
$\Rightarrow \frac{W_{N{a}_{2}C{O}_3}}{106/2}\times 1000=100\times 0.2$
$\therefore W_{N{a}_{2}C{O}_3}=1.06g$

For 90 g of pure $N{a}_{2}C{O}_3,$ weighed sample = 100 g
For 1.06 g of pure $N{a}_{2}C{O}_3,$ weighed sample:
$=\frac{100}{90}\times 1.06g$
$=1.178g$