A) \[{{100.052}^{o}}C\]
B) \[{{101.052}^{o}}C\]
C) \[{{120.52}^{o}}C\]
D) None of these
Correct Answer: A
Solution :
We know that, \[\Delta {{T}_{b}}=\frac{{{K}_{b}}\times {{w}_{B}}\times 1000}{{{M}_{B}}\times {{w}_{A}}}\] \[{{K}_{b}}=0.52K\,\,{{m}^{-1}},\,\,{{w}_{B}}=0.6\,\,g,\,\,{{M}_{b}}=60\] \[{{w}_{A}}=100\,\,g\] \[\therefore \] \[\Delta {{T}_{b}}=\frac{0.52\times 0.6\times 1000}{60\times 1000}={{0.052}^{o}}C\] Thus boiling point of the solution \[=100+0.052={{100.052}^{o}}C\]
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