Question

The concentration of $Ca{\left(HC{O}_3\right)}_2$ in a sample of hard water is $486$ ppm. The density of water sample is $1.0$ g/ml. The molarity of the solution is :

• A. $3.0\times {10}^{-3}M$
• B. $5.0\times {10}^{-3}M$
• C. $2.0\times {10}^{-3}M$
• D. $6.0\times {10}^{-3}M$

Answer 1

Answer: (A)

$3.0\times {10}^{-3}M$

The density of water sample is $1.0$ g/ml.
Thus, $1$ L (or $1000$ mL) of water sample weighs $1$ kg (or $1000$ g).
$486$ ppm corresponds to $486$ g of calcium bicarbonate in $1000$ g or $1000$ L of water sample.
Thus, $1$ L of water sample contains $0.486$ g of calcium bicarbonate.
The molar mass of calcium bicarbonate is $162.11$ g/mol.
The molarity of the solution $=\frac{Massofcalciumbicarbonate}{Molarmassofcalciumbicarbonate\times Volumeofwatersample}=\frac{0.486}{162.11\times 1}=3.0\times {10}^{-3}M$