Question

Calculate the temporary and permanent hardness of water sample having the following the following constituents per litre:
$$ Ca(HCO_{3})_{2} = 162\, mg, MgCl_{2} = 95 =\, mg, $$
$$ NaCl = 585\, mg, Mg(HCO_{3})_{2} = 73\, mg, $$
$$ CaSO_{4} = 136\, mg $$

Answer 1

Correct option is C. 150 ppm, 200 ppm
mole of $$ Ca(HCO_{3})_{2} = \dfrac{162\times 10^{-3}mg}{162\,g/mole} = 1\times 10^{-3}\,moles $$

Mole of $$ Ca(SO_{4}) = \dfrac{136\times 10^{-3}g}{136\,g/mole} = 1\times 10^{3}\,mole $$

Total mole of $$ Ca = 2\times 10^{-3}\,mole $$

mass of $$ CaCO_{3} = 2\times 10^{-3}\times 100 = 0.2\,g $$

$$ \therefore $$ ppm (permanent hardness) $$ = \dfrac{6.2}{1000}\times 10^{6} = 200\,ppm $$

Mole of $$ MgCl_{2} = \dfrac{95\times 10^{-3}}{95} = 1\times 10^{-3}\,mole $$

Mole Mg $$ (HCg)_{2} = \dfrac{73\times 10^{-3}}{146} = 5\times 10^{-4}\,mole $$
mole Mg $$ = 1.5\times 10^{-4}\,mole $$

Mole g $$ CaCO_{3} $$ (In terms of mg) $$ = 1.5\times 10^{-3} $$

mass $$ = 1.5\times 10^{-3} = 0.150\,g $$

ppm (temporary hardness) $$ = \dfrac{0.150}{100}\times 106 = 150\,ppm $$

Hence, the correct option is $$\text{C}$$