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) \[+\Delta {{T}_{b}}\] (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 \[100%\] ionisation [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] \[mi\,\,(\text{urea)}=0.015\] [d] \[mi\,\,(glu\cos e)=0.015\]You need to login to perform this action.
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