A) 100 \[c{{m}^{3}}\]
B) 150 \[c{{m}^{3}}\]
C) 250 \[c{{m}^{3}}\]
D) 200 \[c{{m}^{3}}\]
Correct Answer: D
Solution :
Key Idea: Use the following formula to find V of \[HCl\] used \[V=\frac{w\times 1000}{equivalent\text{ }weight\times N}\] Given, \[w=1.0\,g\] \[N=0.1\] \[\underset{\begin{smallmatrix} 40+16\times 3 \\ +12=100 \end{smallmatrix}}{\mathop{CaC{{O}_{3}}}}\,+2HCl\xrightarrow{{}}CaC{{l}_{2}}+{{H}_{2}}O\] \[\therefore \] 1 mole of \[CaC{{O}_{3}}\] reacts with 2 moles of \[HCl\] \[\therefore \] Equivalent weight \[=\frac{100}{2}=50\] \[\therefore \] \[V=\frac{1\times 1000}{50\times 0.1}\] \[\approx 200\,c{{m}^{3}}\]
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