Freezing point of an aqueous solution is \left ( -0.186 \right )^{\circ}CElevation of boiling point of the same solution isK_{b}= 0.512^{\circ}C,K_{f}= 1.86^{\circ}C find the increase in boiling point.

  • Option 1)

    0.186^{\circ}C

  • Option 2)

    0.0512^{\circ}C

  • Option 3)

    0.092^{\circ}C

  • Option 4)

    0.2372^{\circ}C

 

Answers (1)

As we learnt in

Mathematical Expression -

\Delta T_{b}= K_{b}\: m

Unis of K_{b}=\frac{K-K_{g}}{mole}

\Delta T_{b}= Elevation\: in \: boiling\: point
 

- wherein

K_{b}= Boiling \: point \: elevation \: constant

m= molality

 

 

Mathematical Expression of Depression in Freezing point -

\Delta T_{f}= K_{f}\: m
 

- wherein

m = molarity of solvent 

K_{f} = cryoscopic  constant

    molal depress const

Units = \frac{K-K_{g}}{mole}

 

 \frac{T_{b}}{\Delta T_{f}}=\frac{K_{b}}{K_{f}}

\frac{\Delta T_{b}}{0.186}=\frac{0.512}{1.86}

\Delta T_{b}=0.0512 c^{o}

Correct option is 2.

 


Option 1)

0.186^{\circ}C

This is an incorrect option.

Option 2)

0.0512^{\circ}C

This is the correct option.

Option 3)

0.092^{\circ}C

This is an incorrect option.

Option 4)

0.2372^{\circ}C

This is an incorrect option.

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