# NEET Chemistry Hydrogen Questions Solved

A sample of hard water was found to contain 68 ppm of ${\mathrm{CaSO}}_{4}$ and 19 ppm of ${\mathrm{MgCl}}_{2}$. Its total hardness is –

(A) 70 ppm

(B) 50 ppm

(C) 20 ppm

(D) 100 ppm

(A). 136 ppm ${\mathrm{CaSO}}_{4}$ = 100 ppm ${\mathrm{CaCO}}_{3}$ = 95 ppm ${\mathrm{MgCl}}_{2}$.

Thus 68 ppm ${\mathrm{CaSO}}_{4}$ = 100 × 68 / 136 = 50 ppm ${\mathrm{CaCO}}_{3}$.

19 ppm ${\mathrm{MgCl}}_{2}$ = 100 × 19/95 = 20 ppm ${\mathrm{CaCO}}_{3}$.

Total hardness = 50 + 20 = 70 ppm

Difficulty Level:

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