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 \[{{K}_{f}}=1.86g\,mo{{l}^{-}}\] \[w=45\,g\] \[W=600\,g\] \[m({{C}_{2}}{{H}_{6}}{{O}_{2}})\] \[=2\times 12+6\times 1+2\times 16=62\] \[\therefore \] \[\Delta {{T}_{f-}}=\frac{1.86\times 45\times 1000}{62\times 600}=2.25\,K\]
You need to login to perform this action.
You will be redirected in
3 sec
