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A water sample has the analytical report as under
MgCO3=84 ppm, CaCO3=40 ppm, CaCI2=55.5 ppm, Mg(NO3)2=37 ppm, KCI=10 ppm. Calculate lime & soda required for softening 1 litre of water.
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written 3.4 years ago by
teamques10
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Molecular weight of the impurities are,
- MgCO3$= {24+12+\big(3\times16\big)=84}$
- CaCO3 $= {40+12+\big(3\times16\big)=100}$
- Mg(NO3)2 $= {24+2\big[14+\big(3\times16\big)\big]=148}$
- CaCl2 $= {40+\big(2\times35.5\big)=111}$
Calculations for CaCO3 Equivalents:
| Impurity | Amount in ppm | Molecular Weight | Multiplication Factor $(x) $ | CaCO3 Equivalent in ppm $(y) $ |
|---|---|---|---|---|
| MgCO3 | 84 | 84 | $x = \dfrac{100}{84}$$= {1.19}$ | $y = {Amount\times x}$ $ = {99.96}$ | |
| --- | --- | --- | --- | --- |
| CaCO3 | 40 | 100 | $x = \dfrac{100 } {100}$$= {00 1}$ | $y = {Amount\times x}$$= {40}$ | | Mg(NO3)2 | 37 | 148 | $x = \dfrac{100 } {148}$$= {0.67}$ | $y= {Amount \times x }$$= {25}$ | | CaCl2 | 55.5 | 111 | $x = \dfrac{100}{111}$$= {0.90}$ | $y = {Amount\times x}$$= {50}$ | | KCl | 10 | - | - | - | * KCl do not contribute hardness, so its CaCO3 equivalent is not calculated. * Amount of lime required$= $$\dfrac{74}{100}$${\big[ Temporary \: Ca+ 2\times Temporary \: Mg+ Permanent \:Mg\:\big] }$ $=\dfrac {74}{100}$${\big[40+\big(2 \times 99.96\big)+\big(2 \times 25\big)\big]}$ $= $${214.58\:mg/lit}$ * Amount of Soda required$= $$ \dfrac{106}{100}$${\big[\:Permanent\:Ca+Permanent\:Mg\:\big] }$ $ =$$\dfrac {106}{ 100}$${\big[\:25+50\:\big] }$ $ =$${79.5\:mg/lit}$ * Amount of Lime for softening 1 Litre of Water$ =$$\dfrac{214.58\times1}{10^6}$$= 2.146\times10^{-4}\:kg$ * Amount of Soda for softening 1 Litre of Water$= $$\dfrac{79.5\times1}{10^{6}}$$= 7.95\times10^{-5}\:kg$ * Therefore, for softening one liter of Water, lime required is${2.146\times10^{-4} kg}$ & Soda required is$ {7.95\times10^{-5} kg}$. |
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