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
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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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