Question

# Hardness of water is measured in terms of ppm (parts per million) of $$CaCO_3$$. It is the amount (in g) of $$CaCO_3$$ present in $$10^6 \: g \: H_2O$$. In a sample of water, $$10$$ L required $$0.56$$ g of $$CaO$$ to the remove temporary hardness of $$HCO_3^-$$.$$Ca(HCO_3)_2+CaO\longrightarrow 2CaCO_3+H_2O$$Temporary hardness is :

A
200 ppm CaCO3
B
100 ppm CaCO3
C
50 ppm CaCO3
D
25 ppm CaCO3

Solution

## The correct option is A $$200$$ ppm $$CaCO_3$$$$Ca(HCO_3)_2+CaO\, \, \, \, \, \longrightarrow \, \, \, 2CaCO_3+H_2O$$$$10 \: L\ H_2O\ (= 10^4 \: mL) \: has \: CaCO_3 = 2 \: g$$ $$10^6 \: mL \: (ppm) \: has \:CaCO _3 = 200 \: g$$Thus, $$200 \: ppm \: CaCO_3$$ is the temporary hardness.Chemistry

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