Chemistry, asked by svgurupriya, 5 months ago

A sample of water is found to contain the following analytical data in mgs/lit.Mg(HCO3)2=14.6;MgCl2=9.5;MgSO4=6.0;Ca(HCO3)2=16.2.calculate the temporary and permanent hardness of the sample of water.​

Answers

Answered by nainshusharma18129
2

Answer:

We can calculate the hardness of given water sample as:

(i) Ca(HCO

3

)

2

CaCO3≡CaCO

3

162ppm≡100ppm

∴81ppm≡50ppm

(ii) Mg(HCO

3

)

2

≡CaCO

3

146 ppm ≡ 100 ppm

∴73ppm≡50ppm

(iii) CaSO

4

≡CaCO

3

136ppm≡100ppm

∴68ppm≡50ppm

(iv) MgSO

4

≡CaCO

3

120ppm≡100ppm

∴60ppm≡50ppm

Total degree of hardness of water =50×4=200ppm

Answered by minasharmaminaedu
1

Answer: Temporary hardness = 30.8mg/lit

Permanent hardness = 15.5mg/lit

Explanation:

Temporary hardness causing Impurities = Mg(HCO3 )2, Ca(HCO3 )2

Permanent hardness causing Impurities = MgCl2,MgSO4

Temporary hardness =  14.6+16.2 = 30.8mg/lit

Permanent hardness = 9.5+6 = 15.5mg/lit

Total hardness in water = Temporary hardness + Permanent Hardness

=30.8+15.5

Total hardness in water = 46.3mg/lit

#SPJ3

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