A) \[0.576\text{ }g\]
B) \[0.642\text{ }g\]
C) \[0.444\text{ }g\]
D) \[0.342\text{ }g\]
Correct Answer: C
Solution :
\[\underset{162\,g}{\mathop{Ca{{(HC{{O}_{3}})}_{2}}}}\,+\underset{74\,g}{\mathop{Ca{{(OH)}_{2}}}}\,\xrightarrow{O}2CaC{{O}_{3}}+2{{H}_{2}}O\] 100 mL of pond water has \[Ca{{(HC{{O}_{3}})}_{2}}\] \[=1.62\,mg=1.62\times {{10}^{-3}}g\] \[60,000\,mL\] has \[Ca{{(HC{{O}_{3}})}_{2}}=\frac{1.62\times {{10}^{-3}}}{100}\times 60000\] \[162\,g\]of \[Ca{{(HC{{O}_{3}})}_{2}}\] requires \[Ca{{(OH)}_{2}}=74\,g\] \[0.972\text{ }g\] of \[Ca{{(HC{{O}_{3}})}_{2}}\] requires \[Ca{{(OH)}_{2}}\] \[=\frac{74}{162}\times 0.972=0.444g\]
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