Chemistry, asked by sakharemitesh, 2 days ago

Calculate total, temporary hardness of water sample containing following dissolved salts in mg/L: Mg(HCO3)2 = 29.2 , Ca(HCO3)2 = 32.4, MgSO4= 6, MgCl2 = 9.5 mg/L, Ca(NO3)2= 32.8, CaCl2 = 38.8 , Al2(SO4)2 = 5.7​

Answers

Answered by taniya4429
1

Answer:

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

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