A) \[307.3{}^\circ C\]
B) 307.3 K
C) 34 K
D) \[-34{}^\circ C\]
Correct Answer: B
Solution :
[b] \[\pi =\frac{nRT}{V}\] or \[T=\frac{\pi V}{nR}\] 10% (w/v) aqueous solution of glucose means 10 g glucose is present in 100 mL of \[{{H}_{2}}O\] \[=\frac{14\times 100\times {{10}^{-3}}\times 180}{0.082\times 10}=307.3K\]You need to login to perform this action.
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