A) 2.79 torr
B) 15.7 torr
C) 24.5 torr
D) 32.0 torr
Correct Answer: C
Solution :
Moles of glucose \[=\frac{20.0\,g\,glu\cos e}{180\,g\text{/}mol}=0.111\,\,mol\] Moles of water \[=\frac{70.0\,g\,{{H}_{2}}O}{18.0g/mol}=3.89\,mol\,{{H}_{2}}O\] \[x({{H}_{2}}O)=\frac{3.89}{0.11+3.89}=0.972\] \[p=p{}^\circ {{x}_{{{H}_{2}}O}}=(25.21\,\,\text{torr})\,(0.972)=24.5\,\,\text{torr}\]You need to login to perform this action.
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