The hardness of water sample containing 0.002 mole of magnesium sulphate dissolved in a litre of water is expressed as

Question

Tardigrade Admin · Accepted Answer

The hardness of water sample containing. 0.002 mole of

M g S O_{4}

dissolved in 1 L of water. Number of moles

= \frac{ma ss}{m o l ec u l a r ma ss}

0.02 = \frac{ma ss}{120}

ma ss = 240 \times 10^{- 3} g

ie,

240 \times 10^{- 3} g

mass of

M g S O_{4}

in 1 L of water.

∴

10^{3}

g of

H_{2} O

contains = 0.240 g of

M g S O_{4}

∵

10^{6} g

of

H_{2} O

contains

= \frac{0.240 \times 10 ^{6}}{10 ^{3}} g

of

M g S O_{4}

= 0.240 \times 10^{3} g

of

M g S O_{4}

∵

10^{6}

g of water contains = 240 g of

M g S O_{4}

120 g

M g S O_{4} \equiv 100 g

of

C a C O_{3}

240 g of

M g S O_{4}

= \frac{100 \times 240}{120} = 200 g

of

C a C O_{3}

Hence, hardness of

H_{2} O = 200

ppm

Q. The hardness of water sample containing $0.002$ mole of magnesium sulphate dissolved in a litre of water is expressed as