A) \[15.78\text{ }mL\]
B) \[157.8\text{ }mL\]
C) \[198.4\text{ }mL\]
D) \[295.5\text{ }mL\]
Correct Answer: B
Solution :
Suppose the molecules of \[N{{a}_{2}}C{{O}_{3}}\] and \[NaHC{{O}_{3}}\] in a mixture are a. milli-equivalent of \[N{{a}_{2}}C{{O}_{3}}\] + milli-equivalent of \[NaHC{{O}_{3}}\] milli-equiualent of \[HCl\] \[{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}=NV\] \[a\times 2\times 1000+a\times 1\times 1000=0.1V\] \[3a={{10}^{-4}}V\] ?..(i) [\[\because \]N = basicity/acidity \[\times M\]] wt. of \[N{{a}_{2}}C{{O}_{3}}+wt.\] of \[NaHC{{O}_{3}}=1g\] (\[\because \] wt. of mixture \[=1g\]) \[\Rightarrow \] \[a\times 106+a\times 84=1\] \[a=5.26\times {{10}^{-3}}\] ??(ii) From Eqs. (i) and (ii) we have \[3\times 5.26\times {{10}^{-3}}={{10}^{-4}}V\] \[V=157.8mL\]
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