If hardness is 100 ppm CaCO3, the amount of Na2CO3 required to soften 10 L of hard water is 1.06 g.
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B
If hardness is 100 ppm CaCO3, the amount of Na2CO3 required to soften 10 L of hard water is 10.6 g.
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C
If hardness is 420 ppm MgCO3, the amount of Na2CO3 required to soften 10 L of hard water is 53.0 g.
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D
If hardness is 420 ppm MgCO3, the amount of Na2CO3 required to soften 10 L of hard water is 5.3 g.
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Solution
The correct options are A If hardness is 100 ppm CaCO3, the amount of Na2CO3 required to soften 10 L of hard water is 1.06 g. D If hardness is 420 ppm MgCO3, the amount of Na2CO3 required to soften 10 L of hard water is 5.3 g. Molecular weight of CaCO3=40+12+3×16=100 g/mol Molecular weight of Na2CO3=46+60=106 g/mol
i. CaSO4xmol+Na2CO3xmol→CaCO3xmol+Na2SO4
ii. CaCl2ymol+Na2CO3ymol→CaCO3ymol+2NaCl
iii. 100 ppm CaCO3≡100 g CaCO3 in 106 mL =100100 mol in 106 mL
Moles of Na2CO3 required = Moles of CaCO3 ≡100100 mol in 106 mL ≡1 mol in 106 mL ≡1106×10×103 mL per 10 L =1×10−2 mol in 10 L
Therefore, moles of Na2CO3 required =1×10−2 mol in 10 L =1×10−2×106 g per 10 L =1.06 g
Similarly, for 100 ppm MgCO3 (Molecular weight of MgCO3=24+60=84 g/mol),
iv. MgSO4xmol+Na2CO3xmol→MgCO3xmol+Na2SO4
v. MgCl2ymol+Na2CO3ymol→MgCO3ymol+2NaCl
vi. 420 ppm MgCO3≡420 g MgCO3 in 106 mL ≡42084=5 mol in 106 mL
Moles of Na2CO3 required ≡ moles of MgCO3 ≡5 mol in 106 mL ≡5106×10×103 mL per 10 L ≡5×10−2 mol in 10 L
Moles of Na2CO3 required ≡5×10−2 mol per 10 L ≡5×10−2× per 10 L =5.3 g per 10 L