JEE PYQ: Hydrogen Question 19
Question 19 - 2019 (08 Apr Shift 1)
100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of $\text{CaCO}_3$ is: (molar mass of calcium bicarbonate is 162 g mol$^{-1}$ and magnesium bicarbonate is 146 g mol$^{-1}$)
(a) 5,000 ppm (b) 1,000 ppm (c) 100 ppm (d) 10,000 ppm
Show Answer
Answer: (d)
Solution
Moles of $\text{Ca(HCO}_3)_2 = 0.81/162 = 0.005$. Moles of $\text{Mg(HCO}_3)_2 = 0.73/146 = 0.005$. Hardness $= \frac{(0.005+0.005)\times100}{100}\times10^6 = 10,000$ ppm.
BETA
