Question

$6.2$g of a sample containing $N{a}_{2}C{O}_3$. $NaHC{O}_3$, and non volatile inert impurity on gentle heating losses $5$ percent of its weight. Residue is dissolved in water and $100$mL solution is formed. Its $10$mL portion requires $7.5$mL of $0.2$M aquous solution of $BaC{l}_2$ for complete precipitation. Determine weight (in gram) of $N{a}_{2}C{O}_3$ in the original sample.

• A. $1.59$
• B. $1.06$
• C. $0.53$
• D. None of these

Answer 1

The correct option is B $1.06$

Loss in weight of sample $=\frac{5}{100}\times 6.2=0.31gm$

$2NaHC{O}_3\to N{a}_{2}C{O}_3+C{O}_2+H_{2}O$

According to this reaction, 2 moles of $NaHC{O}_3$ loses 1 mole each of water and carbon dioxide.

(As the impurity is non-volatile, only water and $C{O}_2$ will be lost)

$⟹$ Weight lost by $2molesNaHC{O}_3=18+44=62gm$

$⟹$ Weight lost by $0.01moleNaHC{O}_3=\frac{62}{2}\times 0.01=0.31gm$

Also, 2 moles of $NaHC{O}_3$ gives 1 mole of $N{a}_{2}C{O}_3$

$⟹$ 0.01 moles of $NaHC{O}_3$ gives$\frac{1}{2}\times 0.01=0.005molesofN{a}_{2}C{O}_3$

No of moles of carbonate reacted with given amount of $BaC{l}_2=(\frac{7.5}{1000}\times 0.2\times \frac{100}{10})=0.015$

This is the sum of carbonate already present in the original sample and the carbonate formed due to heating.

$⟹$Moles of carbonates in original sample$=0.015-0.005=0.01$

Molar mass of $N{a}_{2}C{O}_3=106gm$

$⟹$Mass o​f $N{a}_{2}C{O}_3$ in original sample$=0.01\times 106=1.06gm$

So, the correct answer is option (B)