50.0 mL of hard water requires 12.4 mL of 0.0100 M EDTA (1 : 1 complexation with Ca^2+ and Mg^2+). The total hardness in ppm of CaCO3 (molar mass 100) is

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JEE Main levelSingle correct optionDifficulty 2 of 3eq097

$50.0 mL$ of hard water requires $12.4 mL$ of $0.0100 M$ EDTA (1 : 1 complexation with $Ca X^{2 +}$ and $Mg X^{2 +}$ ). The total hardness in ppm of $CaCO X_{3}$ (molar mass $100$ ) is

A248 ppm
B496 ppm
C124 ppm
D24.8 ppm

Answer: A

mol EDTA = 0.0124 x 0.0100 = 1.24e-4 = mol hardness ions. Per litre: 1.24e-4/0.0500 = 2.48e-03 mol. Mass as CaCO3 = 2.48e-03 x 100 g = 0.248 g/L = 248 mg/L = 248 ppm.

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50.0 mL of hard water requires 12.4 mL of 0.0100 M EDTA (1 : 1 complexation with CaX2+ and MgX2+). The total hardness in ppm of CaCOX3​ (molar mass 100) is

Answer: A

More questions on Back titration and hardness of water

$50.0 mL$ of hard water requires $12.4 mL$ of $0.0100 M$ EDTA (1 : 1 complexation with $Ca X^{2 +}$ and $Mg X^{2 +}$ ). The total hardness in ppm of $CaCO X_{3}$ (molar mass $100$ ) is