A) 50
B) 25
C) 75
D) 100
Correct Answer: A
Solution :
Equivalent weight of metal carbonate = 20 + 30 = 50 2.5 g of metal carbonate\[=\frac{2.5}{50}=0.05\,\,\text{eq}\text{.}\] Number of equivalent of \[{{H}_{2}}S{{O}_{4}}\] would have reacted = 0.05. Number of equivalent of \[{{H}_{2}}S{{O}_{4}}\]taken \[=\frac{100\times 1}{1000}=0.1\] Number of equivalent of\[{{H}_{2}}S{{O}_{4}}\]remains unreacted = 0.1 - 0.05 = 0.05 eq. \[\therefore \] Number of equivalent of alkali consumed = 0.05 eq. milli eq. = Normality\[\times \]Volume in mL. \[\therefore \]\[1.0\times V=0.05\times 1000\] \[V=\frac{0.05\times 1000}{1.0}=50\,mL\]You need to login to perform this action.
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