A) \[180\text{ }g\]
B) \[54.2\text{ }g\]
C) \[86\text{ }g\]
D) \[\text{111 }g\]
Correct Answer: B
Solution :
\[n=\frac{\pi V}{RT}\] \[=\frac{7.65\times 1}{0.0821\times 310}=0.301\,mol\] Weight of glucose used \[=0.301\times 180g=54.2g\]You need to login to perform this action.
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