A) \[0.186\]
B) \[0.512\]
C) \[\frac{0.512}{1.86}\]
D) \[0.0512\]
Correct Answer: D
Solution :
\[\Delta {{T}_{f}}=k_{f}^{'}\times m\] \[0.186=1.86\times m\] \[m=\frac{0.186}{1.86}=0.1mol\] \[\therefore \] \[\Delta {{T}_{b}}={{k}_{b}}\times molality\] \[\Delta {{T}_{b}}=0.512\times 0.1\] \[\Delta {{T}_{b}}=0.0512\]You need to login to perform this action.
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