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Question

The concentration of $$\displaystyle Ca\left ( HCO_{3} \right )_{2}$$ 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
3.0×103M
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B
5.0×103M
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C
2.0×103M
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D
6.0×103M
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Solution

The correct option is D $$3.0\times 10^{-3} M$$
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 $$= \dfrac {Mass \ of \ calcium\  bicarbonate} {Molar \ mass\  of\  calcium\  bicarbonate  \times  Volume\  of\  water \ sample} = \dfrac {0.486} {162.11 \times 1} =3.0\times 10^{-3} M$$

Chemistry

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