Question

50 mL of sample of hard water gave good lather with 6mL of standard soap solution (1 mL soap solutions = 1 mg CaCO${}_3$). If the hardness is only due to $Mg(HC{O}_3{)}_2$, the weight of milk of lime required to remove the hardness completely from 100 kg of that sample of water is:

• A. 17.8 g
• B. 8.9 g
• C. 178 g
• D. 89 g

Answer 1

The correct option is B 8.9 g

50 ml hard water $=$ 6 mg $CaC{O}_3$
Assume the density of hard water to be 1 g/mol
50 ml hard water $=$ 50 g or 0.050 kg
(because 1 kg $=$ 1000 g)

Hence, 0.050 kg hard water $=$ 6 mg $CaC{O}_3$
100 kg hard water $=\frac{6mg}{0.050kg}\times 100kg=12000mg=12$ g $CaC{O}_3$
(because 1 g $=$ 1000 mg)
The molecular weights of $Ca(OH{)}_2$ and $CaC{O}_3$ are 74 g/mol and 100 g/mol respectively.

74 g $Ca(OH{)}_2$ $=$ 100 g $CaC{O}_3$

12 g $CaC{O}_3$ $=\frac{74}{100}\times 12=8.9$ g $Ca(OH{)}_2$
Note: Milk of lime is $Ca(OH{)}_2$.