Question

The weight of sodium carbonate required to prepare 500 mL of semi-normal solution is

• A. 8.833 g
• B. 26.5 g
• C. 53 g
• D. 6.125 g

Answer 1

The correct option is A 8.833 g

Semi - normal mean

Normality $=0.5N$

Now normality $=\frac{no.ofmoles\times nfactor}{Volumeofsolution}$

$n-$ factor of sodium Carbonate $=3$

$\{N{a}_{2}C{O}_3⟶2N{a}^{+}+C{O}_3^{2-}\}$

Volume of solution $=500ml$

$=0.5L$

Now,

Normality $=0.5N$

$\Rightarrow \frac{no.ofmoles\times 3}{0.5}=0.5$

$\Rightarrow$ no. of moles $=\frac{0.5\times 0.5}{3}=\frac{0.25}{3}$

weight = molar mass $\times$ no. of moles

$=(23\times 2+12+3\times 16)\times \frac{0.25}{3}$

$=(46+12+48)\times \frac{0.25}{3}$

$=106\times \frac{0.25}{3}=\frac{53}{6}$

$=8.833g$