A water sample analysis data is given below- - Ion Concentration mg-l Atomic Weight- Ca2- 60 40- Mg2- 30 24-31- HCO3- 400 61- -The carbonate hardness expressed as mg-L- of CaCO3- up to one decimal place for the water sample is

Total hardness = [ ca^++(mg/l)×50(40/2)+Mg^++(mg/l)×50(24.31/2) ] = [ 60×5020+30×5012.155 ] =273.4 mg/l as CaCO_3 Alkanity = [ HCO_3^ (mg/l)×5061 ] =400×5061 =3 ...

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Standard XII

Chemistry

Concentration Terms

A water sampl...

Question

A water sample analysis data is given below:

$\begin{matrix} Ion & Concentration m g / l & Atomic Weight C a^{2 +} & 60 & 40 M g^{2 +} & 30 & 24.31 H C O_{3}^{-} & 400 & 61 \end{matrix}$

The carbonate hardness (expressed as mg/L, of $C a C O_{3}$ , up to one decimal place) for the water sample is
273.4

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Solution

The correct option is A 273.4
Total hardness
$= [c a^{+ +} (m g / l) \times \frac{50}{(40 / 2)} + M g^{+ +} (m g / l) \times \frac{50}{(24.31 / 2)}]$
$= [60 \times \frac{50}{20} + 30 \times \frac{50}{12.155}]$
$= 273.4 m g / l$ as $C a C O_{3}$
Alkanity $= [H C O_{3}^{-} (m g / l) \times \frac{50}{61}]$
$= 400 \times \frac{50}{61}$
$= 327.87 m g / l$ as $C a C O_{3}$
Carbonate hardness $=$ Minimum (total hardness, alkanity)
$= 273.4 m g / l$ as $C a C O_{3}$ .

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A water sample analysis data is given below: IonConcentration mg/lAtomic WeightCa2+60 40Mg2+3024.31HCO−3400 61 The carbonate hardness (expressed as mg/L, of CaCO3, up to one decimal place) for the water sample is273.4

The correct option is A 273.4Total hardness =[ca++(mg/l)×50(40/2)+Mg++(mg/l)×50(24.31/2)] =[60×5020+30×5012.155] =273.4mg/l as CaCO3 Alkanity =[HCO−3(mg/l)×5061] =400×5061 =327.87mg/l as CaCO3 Carbonate hardness = Minimum (total hardness, alkanity) =273.4mg/l as CaCO3.