A) \[{{100.01}^{o}}C\]
B) \[{{100.15}^{o}}C\]
C) \[{{100.23}^{o}}C\]
D) \[{{100.32}^{o}}C\]
Correct Answer: B
Solution :
\[p={{p}^{o}}{{x}_{1}}\] \[99.50=100\frac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}\] \[\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{1}{191}\] \[\Delta {{T}_{b}}={{k}_{b}}m={{K}_{b}}\frac{{{n}_{2}}}{{{n}_{1}}}\times \frac{1000}{{{m}_{1}}}\] \[=0.52\times \frac{1}{191}\times \frac{1000}{18}=0.15\] \[{{T}_{b}}={{10015}^{o}}C\]
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