Question

A sample of hard water is found to contain 40mg of $C{a}^{2+}$ ions per litre. The amount of washing soda $(N{a}_{2}C{O}_3)$ required to soften 5.0 L of the sample would be

• A. 1.06 g
• B. 5.3 g
• C. 53 mg
• D. 530 mg

Answer 1

Answer: (D)

The equation is
$C{a}^{2+}+N{a}_{2}C{O}_3\to CaC{O}_3+2N{a}^{+}$
Total $C{a}^{2+}$ in $5L$ of hard water $=5\times 40=200mg$
Now from equation, $40gC{a}^{2+}$ require washing sogd $=106g$
$\therefore 200mg$ of $C{a}^{2+}$ will require washing soda
$=\frac{106}{40}\times 200\times 10^{-3}g$
$=0.530g$ or $530mg$