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Question

One litre of a sample of hard water contains $$55.5$$ mg of $$CaCl_2$$ and $$4.75$$ mg of $$MgCl_2$$. The total hardness in terms of ppm of $$CaCO_3$$ is :


A
5 ppm
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B
10 ppm
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C
20 ppm
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D
none of the above
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Solution

The correct option is B $$10$$ ppm
$$55.5$$ g $$CaCl_2\equiv 50 $$ g $$CaCO_3$$
$$55.5$$ mg $$CaCl_2\equiv 50\times 10^{-3}$$ g $$CaCO_3$$
$$47.5$$ g $$MgCl_2\equiv 50$$ g $$CaCO_3$$
$$47.5 $$ mg $$MgCl_2\equiv 50\times 10^{-3} $$ g $$CaCO_3$$
Total $$CaCO_3=(50+50)\times 10^{-3}=10^{-2}$$ g/L
Hardness in terms of ppm $$=\dfrac {10^{-2}}{1000}\times 10^6$$ mL $$=10$$ ppm

Chemistry

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