JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    The expression to compute molar mass of a solute from the elevation of boiling point of a solvent is (where the various sybols have their usual meanings)

    A) \[{{M}_{2}}=\frac{{{K}_{b}}}{\Delta {{T}_{b}}}\frac{{{m}_{1}}}{{{m}_{2}}}\]          

    B) \[{{M}_{2}}=\frac{\Delta {{T}_{b}}}{{{K}_{b}}}\frac{{{m}_{2}}}{{{m}_{1}}}\]

    C) \[{{M}_{2}}=\frac{{{K}_{b}}}{\Delta {{T}_{b}}}\frac{{{m}_{2}}}{{{m}_{1}}}\]          

    D) \[{{M}_{2}}=\frac{\Delta {{T}_{b}}}{{{K}_{b}}}\frac{{{m}_{1}}}{{{m}_{2}}}\]

    Correct Answer: C

    Solution :

    \[\Delta {{T}_{b}}={{K}_{b}}\times m={{K}_{b}}\times \frac{{{m}_{2}}}{{{M}_{2}}}\times \frac{1}{{{m}_{1}}}\]where \[{{m}_{2}}\]is mass of solute \[{{m}_{1}}\] is mass of solvent \[{{M}_{2}}\] molar mass of solute\[{{M}_{2}}=\frac{{{K}_{b}}\times {{m}_{2}}}{\Delta {{T}_{b}}{{m}_{1}}}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner