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  • Can someone help me with this, The freezing point of a solution prepared from 1.25gm of a non-electrolyte and 20gm of water is 271.9K . If molar depression constant is 1.86Kmole-1 , then molar mass of the solute will be

The freezing point of a solution prepared from 1.25gm of a non-electrolyte and 20gm of water is 271.9K . If molar depression constant is  1.86Kmole-1 , then molar mass of the solute will be

  • Option 1)

    105.7

  • Option 2)

    106.7

  • Option 3)

    115.3

  • Option 4)

    93.9

 

Answers (1)

best_answer

As we learned 

 

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}

 

 

Molar \: mass=\frac{K_{f}\times 1000\times w}{\Delta T_{f}\times W}=\frac{1.86\times 1000\times 1.25}{20\times 1.1}

=105.68 = 105.7

 


Option 1)

105.7

Option 2)

106.7

Option 3)

115.3

Option 4)

93.9

Posted by

Plabita

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