Question

Washing soda $\left(N{a}_{2}C{O}_3.10H_{2}O\right)$ is widely used in softening of hard water. If $1$ L of hard water requires $0.0286$ g of washing soda, the hardness of $CaC{O}_3$ in ppm is :

• A. $10$
• B. $5$
• C. $8$
• D. $6$

Answer 1

Answer: (A)

$10$

Molecular weight of washing soda is $286$ g/mol and of $CaC{O}_3$ is $100$ g/mol.
$CaS{O}_4+N{a}_{2}C{O}_3\to CaC{O}_3+N{a}_{2}S{O}_4$
$0.0286$ g $N{a}_{2}C{O}_3.10H_{2}O=\frac{0.0286}{\frac{286}{2}}\times {10}^3=0.2$ milliequivalents
$0.2$ milliequivalents of $N{a}_{2}C{O}_3=\frac{0.2}{2}$ mmol of $CaC{O}_3$ in $1$ L
$=(0.1\times {10}^3)\times 100$ g in ${10}^3$ mL $=0.01$ g in ${10}^3$ mL

ppm $=\frac{0.01\times {10}^6}{{10}^3}=10$

Hence, the hardness of $CaC{O}_3$ in ppm is $10$ ppm.