A) 20ppm
B) 200 ppm
C) 2000 ppm
D) 120 ppm
E) 240 ppm
Correct Answer: B
Solution :
The hardness of water sample containing. 0.002 mole of\[MgS{{O}_{4}}\]dissolved in 1 L of water. Number of moles \[=\frac{mass}{molecular\text{ }mass}\] \[0.02=\frac{mass}{120}\] \[mass=240\times {{10}^{-3}}g\] ie,\[240\times {{10}^{-3}}g\]mass of\[MgS{{O}_{4}}\]in 1 L of water. \[\therefore \]\[{{10}^{3}}\]g of\[{{H}_{2}}O\]contains = 0.240 g of\[MgS{{O}_{4}}\] \[\because \] \[{{10}^{6}}g\]of\[{{H}_{2}}O\]contains \[=\frac{0.240\times {{10}^{6}}}{{{10}^{3}}}g\]of \[MgS{{O}_{4}}\] \[=0.240\times {{10}^{3}}g\]of\[MgS{{O}_{4}}\] \[\because \]\[{{10}^{6}}\]g of water contains = 240 g of \[MgS{{O}_{4}}\] 120 g \[MgS{{O}_{4}}\equiv 100g\]of\[CaC{{O}_{3}}\] 240 g of\[MgS{{O}_{4}}\] \[=\frac{100\times 240}{120}=200g\]of\[CaC{{O}_{3}}\] Hence, hardness of\[{{H}_{2}}O=200\]ppm
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