A) 880.07 g
B) 899.04 g
C) 866.02 g
D) 868.06 g
Correct Answer: D
Solution :
\[\Delta {{T}_{f}}={{K}_{f}}m\]where m = molality \[273-268=1.86\times \frac{w}{M\times v}\] \[5=1.86\times \frac{w}{32\times 10}\] \[w=\frac{5\times 32\times 10}{1.86}\] \[=860.2\approx 868.06g\]You need to login to perform this action.
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