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:
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$ {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
