A) 7.9 K
B) 2.5 K
C) 6.6 K
D) 2.2 K
Correct Answer: D
Solution :
Depression of freezing point is given as \[\Delta {{T}_{f}}=\frac{{{K}_{f}}\times W\times 1000}{m\times W}\] For water, \[{{K}_{f}}=1,86g\text{ }mo{{l}^{-}}\] \[w=45\text{ }g\] \[W=600\text{ }g\] \[m({{C}_{2}}{{H}_{6}}{{O}_{2}})\] \[=2\times 12+6\times 1+2\times 16=62\] \[\therefore \]\[\Delta {{T}_{f-}}=\frac{1.86\times 45\times 1000}{62\times 600}=2.25\,K\]
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