A) 0.84
B) 8.4
C) 33.6
D) 16.8
Correct Answer: B
Solution :
\[2NaHC{{O}_{3}}+{{(COOH)}_{2}}\xrightarrow[{}]{{}}\]\[{{(COONa)}_{2}}+2{{H}_{2}}O+2C{{O}_{2}}\]\[0.25mL\] \[T=298.15K,P=1bar\] \[n=\frac{0.25}{25000}={{10}^{-5}}\] No. of moles \[(n)=\frac{Weight}{Molecular\,weight}\] \[w=84\times {{10}^{-5}}g=\frac{84\times {{10}^{-5}}}{{{10}^{-2}}}\times 100=8.4%\]You need to login to perform this action.
You will be redirected in
3 sec