100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO3 is: (molar mass of calcium bicarbonate is 162 g mol-1 and magnesium bicarbonate is 146 g mol-1)

Question

100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO3 is: (molar mass of calcium bicarbonate is 162 g mol-1 and magnesium bicarbonate is 146 g mol-1) (A) 1,000 ppm (B) 10,000 ppm (C) 100 pm (D) 5,000 pm

Tardigrade Admin · Accepted Answer

neq . CaCO3 = neq Ca(HCO3)2 + neqMG(HCO3)2 or (W/100) ×2 = (0.81/162) ×2 + (0.73/146) ×2 ∴ w = 1.0 ∴ Hardness = (1.0/100) × 106 = 10000 ppm

$1, 000 pp m$

$10, 000 pp m$

$100 p m$

$5, 000 p m$

$n_{e q} . C a C O_{3} = n_{e q} C a (H C O_{3})_{2} + n_{e q} MG (H C O_{3})_{2}$
or $\frac{W}{100} \times 2 = \frac{0.81}{162} \times 2 + \frac{0.73}{146} \times 2$
$∴ w = 1.0$
$∴$ Hardness $= \frac{1.0}{100} \times 1 0^{6} = 10000 pp m$

Q. 100mL of a water sample contains 0.81g of calcium bicarbonate and 0.73 of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO3​ is: (molar mass of calcium bicarbonate is 162gmol−1 and magnesium bicarbonate is 146gmol−1)

1,000ppm

10,000ppm

100pm

5,000pm

neq​.CaCO3​=neq​Ca(HCO3​)2​+neq​MG(HCO3​)2​ or 100W​×2=1620.81​×2+1460.73​×2 ∴w=1.0 ∴ Hardness =1001.0​×106=10000ppm

$1, 000 pp m$

$10, 000 pp m$

$100 p m$

$5, 000 p m$

$n_{e q} . C a C O_{3} = n_{e q} C a (H C O_{3})_{2} + n_{e q} MG (H C O_{3})_{2}$
or $\frac{W}{100} \times 2 = \frac{0.81}{162} \times 2 + \frac{0.73}{146} \times 2$
$∴ w = 1.0$
$∴$ Hardness $= \frac{1.0}{100} \times 1 0^{6} = 10000 pp m$