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Can someone help me with this, - Solutions - JEE Main-3

Molal depression constant for a solvent is 4.0\:\:K\:kg\:mol^{-1}. The depression in the freezing point of the solvent for 0.03\:\:\:mol\:kg^{-1}\:\:solution\:\:of\:\:K_{2}SO_{4}  is :

( Assume complete dissociation of the electrolyte )

  • Option 1)

    0.18\:K

  • Option 2)

    0.24\:K

  • Option 3)

    0.12\:K

  • Option 4)

    0.36\:K

Answers (1)
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Depression in freezing point -

-

 

 

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}

 

 

Vant Hoff factor (i) -

In case of electrolytes the observed colligative property is different from theoritical colligative property. There ratio is defined by Vant Haff factor

- wherein

i = \frac{observed}{theoritical}

 

 

 

 

K_{2}SO_{4}\rightarrow 2K^{+}+SO_{4}^{2+}

i=2+1=3

\\\Delta T_{f}=K_{f}\times i\times m\\\\\ \Delta T_{f}:=4(K\:\:kg/mol)\times 3 \times 0.03

\\\Delta T_{f}=0.36\:\:K


Option 1)

0.18\:K

Option 2)

0.24\:K

Option 3)

0.12\:K

Option 4)

0.36\:K

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