A) 180 g
B) 54.2 g
C) 86 g
D) 111 g
Correct Answer: B
Solution :
\[n=\frac{\pi V}{RT}=\frac{(7.65atm)(1.00L)}{(0.0821Latm)/molK)(310K)}\]\[=0.301mol\] Weight of glucose used =\[(0.301mol)(180g{{C}_{6}}{{H}_{12}}{{O}_{6}}/mol)=54.2g\]You need to login to perform this action.
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