A) 54.2 g
B) 90 g
C) 12 g
D) 0.30 g
Correct Answer: A
Solution :
\[pV=nRT\] \[n=\frac{pV}{RT}=\frac{(7.65\,atm)(1.00\,L)}{(0.0821\,L\,atm/mol\,K)\,(310\,K)}\] \[= \,\,0.301 mol\] \[\operatorname{Mass} of glucose = \left( 0.301 mol \right)\] \[(180g/mol\,{{C}_{6}}{{H}_{12}}{{O}_{6}})=54.2g\]You need to login to perform this action.
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