Question

$9.5\times {10}^{-3}\%MgC{l}_2$ and $1.36\times {10}^{-4}\%CaS{O}_4$ are present in water. What is the measured hardness of water in terms of $CaC{O}_3$?
(Given : molar mass of $MgC{l}_{2}andCaS{O}_4$ are 95 and 136 g/mol respectively. )

• A. $20ppm$
• B. $56ppm$
• C. $200ppm$
• D. $101ppm$

Answer 1

The correct option is D $101ppm$

$\text{Equivalents of}CaC{O}_3=\text{Equivalents of}MgC{l}_2+\text{Equivalents of}CaS{O}_4$

$\frac{W_{CaC{O}_3}}{100}\times 2=\frac{0.0095}{95}\times 2+\frac{0.000136}{136}\times 2W_{CaC{O}_3}=(1\times {10}^{-4}+1\times {10}^{-6})\times 100$
$\therefore W_{CaC{O}_3}\approx 0.0101g$

$W_{H_{2}O}=100g$

$\text{Hardness}=\frac{\text{Mass of}CaC{O}_3}{\text{Mass of hard water}}\times {10}^{6}ppm$

$\therefore \text{Hardness}=\frac{0.0101}{100}\times {10}^{6}ppm$
$\text{Hardness}=101ppm$