A) \[0.08\,\,M\,\,BaC{{l}_{2}}\]
B) \[0.15\,\,M\,\,KCl\]
C) \[0.10\,\,M\]Glucose
D) \[0.06\,\,M\,\,Ca{{(N{{O}_{3}})}_{2}}\]
Correct Answer: B
Solution :
\[\because \] \[\Delta {{T}_{b}}=\frac{1000\cdot {{K}_{b}}\cdot w}{m\cdot W}\times i\] \[\therefore \] \[\Delta {{T}_{b}}\propto i\cdot \frac{w}{m\cdot \frac{W}{1000}}(\because \,\,{{K}_{b}}=\text{constant})\] or \[\Delta {{T}_{b}}\propto i\cdot M\] (\[\because \]Molarity\[=\frac{w}{m\cdot \frac{W}{1000}}\]and assuming, Molarity,\[M=\]molarity,\[m\]) Now, for the given solutions, (a) For\[0.08M\,\,BaC{{l}_{2}}\,\,i\cdot M=3\times 0.08=0.24\] (b) For\[0.15\,\,M\,\,KCl,\,\,i\cdot M=2\times 0.15=0.30\] (c) For\[0.10\,\,M\]glucose,\[i\cdot M=1\times 0.10=0.10\] (d) For\[0.06\,\,M\,\,Ca{{(N{{O}_{3}})}_{2}},\,\,i\cdot M=3\times 0.06\] \[=0.18\] Hence,\[\Delta {{T}_{b}}\]is maximum for\[0.15\,\,M\,\,KCl\]solution.
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