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Question

Chemistry Question on Types of Solutions

The hardness of water sample containing 0.0020.002 mole of magnesium sulphate dissolved in a litre of water is expressed as

A

20ppm20\, ppm

B

200ppm200\, ppm

C

2000ppm2000\, ppm

D

120ppm120\, ppm

Answer

200ppm200\, ppm

Explanation

Solution

The hardness of water sample containing. 0.002 mole of MgSO4MgS{{O}_{4}} dissolved in 1 L of water. Number of moles =massmolecular mass=\frac{mass}{molecular\text{ }mass} 0.02=mass1200.02=\frac{mass}{120} mass=240×103gmass=240\times {{10}^{-3}}g ie, 240×103g240\times {{10}^{-3}}g mass of MgSO4MgS{{O}_{4}} in 1 L of water. \therefore 103{{10}^{3}} g of H2O{{H}_{2}}O contains = 0.240 g of MgSO4MgS{{O}_{4}} \because 106g{{10}^{6}}g of H2O{{H}_{2}}O contains =0.240×106103g=\frac{0.240\times {{10}^{6}}}{{{10}^{3}}}g of MgSO4MgS{{O}_{4}} =0.240×103g=0.240\times {{10}^{3}}g of MgSO4MgS{{O}_{4}} \because 106{{10}^{6}} g of water contains = 240 g of MgSO4MgS{{O}_{4}} 120 g MgSO4100gMgS{{O}_{4}}\equiv 100g of CaCO3CaC{{O}_{3}} 240 g of MgSO4MgS{{O}_{4}} =100×240120=200g=\frac{100\times 240}{120}=200g of CaCO3CaC{{O}_{3}} Hence, hardness of H2O=200{{H}_{2}}O=200 ppm