Question

0.7g of $N{a}_2$C$O_3$. x $H_{2}O$ is dissolved in 100mL of water, 20mL of which required 19.8mL of 0.1 0.1N HCl. The value of x is :

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

2

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

Number of mole of $HCl=19.8\times 0.1\times {10}^{-3}$

$=1.98\times {10}^{-3}$

$2molHCl$ required $1molN{a}_{2}C{O}_3$

$1.98\times {10}^{-3}$ _________ $\frac{1}{2}\times 1.98\times {10}^{-3}$

$0.99\times {10}^{-3}molN{a}_{2}C{O}_3$

Now, $0.7g$ of $N{a}_{2}C{O}_3.x{H}_{2}O$ in $100ml$

Molarity $=\frac{0.7\times 1000}{\frac{(46+12+48+18x)}{100ml}}=\frac{0.7\times 10}{(106+18x)}$

$=\frac{7}{106+18x}$

Number of mol of $N{a}_{2}C{O}_3$ in $20ml.$

$\frac{7}{106+18x}\times 20\times {10}^{-3}=0.99\times {10}^{-3}$

$\frac{140}{0.99}=106+18x$

$x\simeq 2$