A) \[0.01\text{ }M\text{ }N{{a}_{2}}S{{O}_{4}}\]
B) \[0.01\text{ }M\text{ }KN{{O}_{3}}\]
C) \[0.015\text{ }M\] urea
D) 0.015 M glucose
Correct Answer: A
Solution :
Boiling point\[={{T}_{o}}\](Solvent)\[+{{T}_{b}}\](Elevation in bp) \[\Delta {{T}_{b}}=mi{{K}_{b}}\] where, m is the molality (\[\approx \]Molarity M) \[i,\]the van't Hoff factor\[=[1+(y-1)x]\] \[{{K}_{b,}}\]molal elevation constant. Thus,\[\Delta {{T}_{b}}\propto i\] Assume 100% ionisation \[mi(N{{a}_{2}}S{{O}_{4}})=0.01\times 3=0.03\] \[mi(KN{{O}_{3}})=0.01\times 2=0.02\] \[mi(urea)=0.015\] \[mi(glu\cos e)=0.015\]
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