Question

?
Calculate the temporary hardness in terms of calcium carbonate equivalents in a water sample
containing calcium bicarbonate (12.2 mg). Given that at. Wt. of Ca=40 amu, O=16 amu, C=12 amu,
H=1 amu,​

Answer 1

mole of Ca(HCO

3

)

2

=

162g/mole

162×10

−3

mg

=1×10

−3

moles

Mole of Ca(SO

4

)=

136g/mole

136×10

−3

g

=1×10

3

mole

Total mole of Ca=2×10

−3

mole

mass of CaCO

3

=2×10

−3

×100=0.2g

∴ ppm (permanent hardness) =

1000

6.2

×10

6

=200ppm

Mole of MgCl

2

=

95

95×10

−3

=1×10

−3

mole

Mole Mg (HCg)

2

=

146

73×10

−3

=5×10

−4

mole

mole Mg =1.5×10

−4

mole

Mole g CaCO

3

(In terms of mg) =1.5×10

−3

mass =1.5×10

−3

=0.150g

ppm (temporary hardness) =

100

0.150

×106=150ppm

Hence, the correct option is C

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