A) 5
B) 10
C) 15
D) 20
Correct Answer: B
Solution :
\[\underset{\begin{smallmatrix} 40+35.5\times 2 \\ =111\,\times \,{{10}^{3}}\,mg \end{smallmatrix}}{\mathop{CaC{{l}_{2}}}}\,\equiv \underset{\begin{smallmatrix} 40\,+\,12\,+\,48 \\ =\,100\,\,\times \,\,{{10}^{3}}\,mg \end{smallmatrix}}{\mathop{CaC{{O}_{3}}\,}}\,\,\,\,\,\,;\,\,\] \[\underset{\begin{smallmatrix} 24\,+\,32\,+\,64 \\ =120\,\times \,{{10}^{3}}\,mg \end{smallmatrix}}{\mathop{MgS{{O}_{4}}}}\,\,\,\,\,\,\,\,\,\,\,\equiv \,\,\underset{\begin{smallmatrix} 40\,\,+\,\,12\,\,+\,\,48 \\ =\,100\,\,\times \,\,{{10}^{3}}\,mg \end{smallmatrix}}{\mathop{CaC{{O}_{3}}\,}}\,\,\,\,\,\] \[111\times {{10}^{3}}\,\text{mg}\]of \[CaC{{l}_{2}}=100\times {{10}^{3}}\,\text{mg}\,\]of \[CaC{{O}_{3}}\] \[11.1\,\text{mg}\]of \[CaC{{l}_{2}}=\frac{100\times {{10}^{3}}\times 11.1}{111\times {{10}^{3}}}mg\]of \[CaC{{O}_{3}}\] \[=10\,mg\]of \[CaC{{O}_{3}}\] Similarly, \[120\times {{10}^{3}}\text{mg}\]of \[MgS{{O}_{4}}=100\times {{10}^{3}}\,mg\]of \[CaC{{O}_{3}}\] \[12\,mg\]of \[MgS{{O}_{4}}=\frac{100\times {{10}^{3}}}{120\times {{10}^{3}}}\times 12\,mg\]of \[CaC{{O}_{3}}\] \[=10\,mg\]of \[CaC{{O}_{3}}\] In 2 L of water, total weight of \[CaC{{O}_{3}}\] \[=10+10=20\,mg\] \[\therefore \]In 1L of water, total weight of \[CaC{{O}_{3}}\] \[=\frac{20}{2}\,mg\] \[=10\,mg\] \[\therefore \]In \[\text{1}{{\text{0}}^{\text{6}}}\,\text{mg}\]of water, total weight of \[\text{CaC}{{\text{O}}_{3}}=10\,mg\] So, in \[{{10}^{6}}\] part water, hardness of water in terms of \[\text{CaC}{{\text{O}}_{\text{3}}}\text{= 10}\,\text{mg}\]
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