Calculate the temporary and permanent hardness of water sample having the following the following constituents per litre:
$C a (H C O_{3})_{2} = 162 m g, M g C l_{2} = 95 = m g,$
$N a C l = 585 m g, M g (H C O_{3})_{2} = 73 m g,$
$C a S O_{4} = 136 m g$

Question

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Answer 1

mole of $C a (H C O_{3})_{2} = \frac{1 6 2 \times 1 0 ^{- 3} m g}{1 6 2 g / m o l e} = 1 \times 1 0^{- 3} m o l e s$

Mole of $C a (S O_{4}) = \frac{1 3 6 \times 1 0 ^{- 3} g}{1 3 6 g / m o l e} = 1 \times 1 0^{3} m o l e$

Total mole of $C a = 2 \times 1 0^{- 3} m o l e$

mass of $C a C O_{3} = 2 \times 1 0^{- 3} \times 100 = 0.2 g$

$∴$ ppm (permanent hardness) $= \frac{6 . 2}{1 0 0 0} \times 1 0^{6} = 200 p p m$

Mole of $M g C l_{2} = \frac{9 5 \times 1 0 ^{- 3}}{9 5} = 1 \times 1 0^{- 3} m o l e$

Mole Mg $(H C g)_{2} = \frac{7 3 \times 1 0 ^{- 3}}{1 4 6} = 5 \times 1 0^{- 4} m o l e$
mole Mg $= 1.5 \times 1 0^{- 4} m o l e$

Mole g $C a C O_{3}$ (In terms of mg) $= 1.5 \times 1 0^{- 3}$

mass $= 1.5 \times 1 0^{- 3} = 0.150 g$

ppm (temporary hardness) $= \frac{0 . 1 5 0}{1 0 0} \times 106 = 150 p p m$

Hence, the correct option is $C$

Answer 2

mole of

C a (H C O_{3})_{2} = \frac{1 6 2 \times 1 0 ^{- 3} m g}{1 6 2 g / m o l e} = 1 \times 1 0^{- 3} m o l e s

Mole of

C a (S O_{4}) = \frac{1 3 6 \times 1 0 ^{- 3} g}{1 3 6 g / m o l e} = 1 \times 1 0^{3} m o l e

Total mole of

C a = 2 \times 1 0^{- 3} m o l e

mass of

C a C O_{3} = 2 \times 1 0^{- 3} \times 100 = 0.2 g

∴

ppm (permanent hardness)

= \frac{6 . 2}{1 0 0 0} \times 1 0^{6} = 200 p p m

Mole of

M g C l_{2} = \frac{9 5 \times 1 0 ^{- 3}}{9 5} = 1 \times 1 0^{- 3} m o l e

Mole Mg

(H C g)_{2} = \frac{7 3 \times 1 0 ^{- 3}}{1 4 6} = 5 \times 1 0^{- 4} m o l e

mole Mg

= 1.5 \times 1 0^{- 4} m o l e

Mole g

C a C O_{3}

(In terms of mg)

= 1.5 \times 1 0^{- 3}

mass

= 1.5 \times 1 0^{- 3} = 0.150 g

ppm (temporary hardness)

= \frac{0 . 1 5 0}{1 0 0} \times 106 = 150 p p m

Hence, the correct option is

C