JEE Main & Advanced Sample Paper JEE Main Sample Paper-10

  • question_answer
    If 0.1 M\[{{H}_{2}}S{{O}_{4}}(aq)\] solution shows freezing point \[-0.39{}^\circ C\], then what is\[{{K}_{{{a}_{2}}}}\] for\[{{H}_{2}}S{{O}_{4}}\] (Given, molality = molarity and\[{{K}_{f}}_{({{H}_{2}}O)}=1.86\] kg mol-1)

    A)  0.012                   

    B)  0.015

    C)  0.018                   

    D)  2.0

    Correct Answer: A

    Solution :

     Idea This problem is based on concept of V ant Hoff factor and dissociation constant. Students are advised to calculate V ant Hoff factor first followed by calculation of dissociation constant step by step. \[\Delta {{T}_{f}}={{K}_{f}}\times m\times i=0.39=i\times 1.86\times 0.1\] \[i\approx 2.1\] \[\underset{0}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,\xrightarrow{{}}\underset{c}{\mathop{{{H}^{+}}}}\,+\underset{c(1-\alpha )}{\mathop{HSO_{4}^{-}}}\,\] \[\underset{c(1-\alpha )}{\mathop{HSO_{4}^{-}}}\,\underset{c\alpha }{\mathop{{{H}^{+}}}}\,+\underset{c\alpha }{\mathop{SO_{4}^{-}}}\,\] TEST Edge Question involving concept of V ant Hoff factor and dissociation constant is asked in JEE Main generally. The question including calculation of \[p{{K}_{a}}\] may also be asked so students are advised to go through concept of \[p{{K}_{a}}\] also \[p{{K}_{a}}=-\log {{K}_{a}}\]


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