Explanation

(b)
The hardness of water sample containing 0.02 mole of $MgSO_{4}$ dissolved in 1 L of water.
Number of moles = mass/molecular mass
0.002 = mass/120
mass = 0.24 g
0.24 g mass of $MgSO_{4}$ in 1 L of water.
$\therefore 10^{3}\ g\ of\ H_{2}O\ contains = 0.24\ g\ of\ MgSO_{4}$
$\because 10^{6}\ g\ of\ H_{2}O\ contains = \frac{0.24 \times 10^{6}}{10^{3}}\ g\ of\ MgSO_{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.24 \times 10^{3}\ g = 0.24\ g\ of\ MgSO_{4}$
$\ 10^{6}\ g\ of\ water\ contains = 240\ g\ of\ MgSO_{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 120\ g\ MgSO_{4} \equiv 100\ g\ of\ CaCO_{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 240\ g\ of\ MgSO_{4} = \frac{100 \times 240}{120} = 200\ g\ of\ CaCO_{3}$
Hence, hardness of $H_{2}O = 200\ ppm.$