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 CaCO₃ is: (Molar mass of calcium bicarbonate is 162 g.mol^-1 and magnesium bicarbonate is 146 g.mol^-1)

Question

Question

This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 1)

Answer 1

Concept:

The basic formula of Parts per Million is:

\({\rm{PPM}} = \frac{{{\rm{Weight}}}}{{{\rm{Volume}}}} \times {10^6}\)

The degree of hardness and the weight are interrelated. Thus,

\({\rm{Degree\;of\;hardness}} = \frac{{{\rm{\;Weight\;of\;hardness\;causing\;salt\;}}}}{{{\rm{Molecular\;weight}}}} \times 100\)

Calculation:

\(\Rightarrow {\rm{Degree\;of\;hardness}} = \left( {\frac{{0.81}}{{162}} + \frac{{0.73}}{{146}}} \right) \times 100\)

∴ Degree of hardness=1

Now, \({\rm{PPM}} = \frac{1}{{100}} \times {10^6}\)

∴ PPM = 10000 ppm