A) \[0.01\,M\,N{{a}_{2}}S{{O}_{4}}\]
B) \[0.01\,M\,KN{{O}_{3}}\]
C) \[0.015\,M\,\text{urea}\]
D) \[0.015\,M\,\text{glucose}\]
Correct Answer: A
Solution :
Boiling point \[={{T}_{0}}(Solvent)+{{T}_{b}}\] (Elevation in b. p.) \[\Delta {{T}_{b}}=mi{{k}_{b}}\] where, m is the molality ie, the van't Hoff factor (i) \[=[1+(y-1)x]\] \[{{k}_{b}}=\]molal elevation constant. Thus, \[\ \Delta {{T}_{b}}\propto im\] Assume 100% ionization [a]\[mi(N{{a}_{2}}S{{O}_{4}})=0.01\times 3=0.03\] [b] \[mi(KN{{O}_{3}})=0.01\times 2=0.02\] [c] \[mi(glucose)=0.015\]
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