JEE MAIN - Chemistry (2019 - 12th January Morning Slot - No. 12)
The hardness of a water sample (in terms of equivalents of CaCO3) containing 10–3 M CaSO4 is (Molar mass of CaSO4 = 136 g mol–1)
90 ppm
100 ppm
50 ppm
10 ppm
Explanation
nCaSO4 $$ \times $$ Van't hoff factor = nCaCO3 $$ \times $$ Van't hoff factor
$$ \Rightarrow $$ 10-3 $$ \times $$ 2 = nCaCO3 $$ \times $$ 2
$$ \Rightarrow $$ nCaCO3 = 10-3 mol in 1 L
$$ \therefore $$ wCaCO3 = 100 $$ \times $$ 10-3 g CaCO3 in 1 L solution
$$ \therefore $$ hardness in terms of CaCO3
= $${{{w_{CaC{O_3}}}} \over {{w_{Total}}}} \times {10^6}$$
= $${{100 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}$$
= 100 ppm
$$ \Rightarrow $$ 10-3 $$ \times $$ 2 = nCaCO3 $$ \times $$ 2
$$ \Rightarrow $$ nCaCO3 = 10-3 mol in 1 L
$$ \therefore $$ wCaCO3 = 100 $$ \times $$ 10-3 g CaCO3 in 1 L solution
$$ \therefore $$ hardness in terms of CaCO3
= $${{{w_{CaC{O_3}}}} \over {{w_{Total}}}} \times {10^6}$$
= $${{100 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}$$
= 100 ppm
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