JEE Main 12 Jan 2019 Morning Question 13

Question: The hardness of a water sample (in terms of equivalents of $ CaCO _3 $ ) containing $ {10^{-3}}M,CaSO _4 $ is (molar mass of $ ,CaSO _4=136g,mo{l^{-1}} $ ) [JEE Main Online Paper Held On 12-Jan-2019 Morning]

Options:

A) 10 ppm

B) 100 ppm

C) 50 ppm

D) 90 ppm

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ {10^{-3}}M $ means $ {10^{-3}} $ moles of $ CaSO _4 $ present in 1 L of water.

    $ {10^{-3}}moles,CaSO _4=\frac{Mass,of,CaSO _4}{Molar,mass,of,CaSO _4} $

    Mass of $ CaSO _4={10^{-3}}mol\times 136g,mo{l^{-1}} $ or 136 mg i.e., $ \because $ 136 mg of $ CaSO _4 $ present in 1 kg of water
    $ \therefore $ $ 10^{6}g $ of water will have $ 136000mg,CaSO _4 $ $ 136gCaSO _4\equiv 100g,CaCO _3 $ $ 13600mg,CaSO _4\equiv \frac{100}{136}\times \frac{136000}{1000}=100g,CaCO _3 $ Thus, hardness of water =100 ppm