A) 0.80 K
B) 0.40 K
C) 0.60 K
D) 0.20 K
Correct Answer: B
Solution :
| [b] \[\Delta {{T}_{f}}={{k}_{f}}m\] |
| \[=5.12\text{ (}K.kg\text{ }mo{{l}^{1}})\times 0.078\text{ (}mol\text{ }k{{g}^{1}})\] |
| \[=0.399\text{ }K\] |
| \[\approx \,~0.40\text{ }K\] |
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