A) 2
B) 0.30
C) 4
D) 0.20
Correct Answer: C
Solution :
\[\underset{\begin{smallmatrix} 100\,gm \\ (one\,mol.) \end{smallmatrix}}{\mathop{CaC{{O}_{3}}}}\,+\underset{\begin{smallmatrix} 98\,gm \\ (one\,mol.) \end{smallmatrix}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,\to CaS{{O}_{4}}+C{{O}_{2}}+{{H}_{2}}O\] \[\because \]100 gm\[CaC{{O}_{3}}=98\,gm\,{{H}_{2}}S{{O}_{4}}\] \[\therefore \]\[100\,gm\,CaC{{O}_{3}}=\frac{98\times 10}{100}\] \[=9.8\,gm\,of\,{{H}_{2}}S{{O}_{4}}\] \[\therefore \]50 ml of solution have 9.8 gm of\[{{H}_{2}}S{{O}_{4}}\] \[\therefore \]1000 ml of solution have \[=\frac{9.8\times 1000}{50}\] = 196 gm of\[{{H}_{2}}S{{O}_{4}}\] We know that \[N=\frac{W}{equivalent\text{ }weight}\times \frac{1000}{V}\] \[=\frac{196\times 1000}{49\times 1000}=4\] \[\therefore \] \[N=4\]
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