Question

A sample of hard water contains 20 ppm Ca++. How many moles of na2co3 required to remove the hardness of 10 Litre of water?

Answer 1

Answer:

55.5 g CaCl

2

≡50 g CaCO

3

55.5 mg CaCl

2

≡50×10

−3

g CaCO

3

47.5 g MgCl

2

≡50 g CaCO

3

47.5 mg MgCl

2

≡50×10

−3

g CaCO

3

Total CaCO

3

=(50+50)×10

−3

=10

−2

g/L

Hardness in terms of ppm =

1000

10

−2

×10

6

mL =10 ppm

Answer 2

We require $0.005 mol$  $Na_{2} CO_{3}$   to precipitate the $Ca2+$

Explanation:

• Given sample of water for which the hardness is to be determined is 10 L water
• to find the number of moles of any substance.
• Number of moles n , is given by:
• n = given mass/Molar mass
• Given mass should be converted to gm and Molar mass is given in gm/mol.

Therefore you need to precipitate 20 mg×10 = 200 mg $Ca^{+2}$

Equation which is involved i :-

$Ca^{+2} +Na_{2} CO_{3} ----CaCO_{3} + 2Na^{+2}$

⇒ When 1 mol $Ca^{+2}$ requires 1 mol  $Na_{2} CO_{3}$

⇒Molar mass $Ca = 40 g/mol$

⇒When $1 mol$ Ca in $200 mg$ = $0.200 g$

⇒$0.200g / 40 g/mol$ =$0.005 mol Ca$

⇒We require $0.005 mol$  $Na_{2} CO_{3}$   to precipitate the $Ca2+$

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