A) \[0.064\]
B) \[0.045\]
C) \[0.015\]
D) \[0.032\]
Correct Answer: D
Solution :
: \[\underset{1mol}{\mathop{{{(COOH)}_{2}}}}\,+\underset{2moles}{\mathop{2NaOH}}\,\to {{(COONa)}_{2}}+2{{H}_{2}}O\] Mol. mass of \[NaOH=40g\,mo{{l}^{-1}}\] No. of g moles in \[0.064g\]of \[NaOH\] \[=\frac{0.064}{40}=8\times {{10}^{-4}}\] No. of moles of oxalic acid \[=\frac{0.0016}{2}=8\times {{10}^{-4}}\] Volume of solution (in L) \[=\frac{25}{1000}\] Hence, molarity \[=\frac{No.\,of\,moles\,of\,solute}{Volume\,of\,solution\,(in\,L)}\] \[=8\times {{10}^{-4}}\times \frac{1000}{25}=0.032M\]
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