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  • Confused! kindly explain, In equimolar solution of glucose, NaCl and BaCl2the order of osmotic pressure is as follow

In equimolar solution of glucose, NaCl and BaCl2 the order of osmotic pressure is as follow

  • Option 1)

    Glucose> NaCl > BaCl2

  • Option 2)

    NaCl > BaCl2  > Glucose

  • Option 3)

    BaCl2 > NaCl > Glucose

  • Option 4)

    Glucose >BaCl2 > NaCl

 

Answers (1)

best_answer

As we learned

 

Vant Hoff factor for dissociation -

i= 1+(n-1)\alpha

Where

n is the no. of dissociated particles

\alpha = degree of dissociation
 

- wherein

NaC l \: \: \: n = 2

CaCl_{2} \: \: \: n = 3

K_{4}[F(CN_{6})]\: \: \: \: \: n=5

 

 

BaCl_{2}\Rightarrow Ba^{2+}+2Cl^{-}=3\: ion

NaCl\Rightarrow Na^{+}+Cl^{-}=2\: ion

Glucose\Rightarrow No\: ionisation

\therefore BaCl_{2} > NaCl > Glucose

 


Option 1)

Glucose> NaCl > BaCl2

Option 2)

NaCl > BaCl2  > Glucose

Option 3)

BaCl2 > NaCl > Glucose

Option 4)

Glucose >BaCl2 > NaCl

Posted by

Plabita

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