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The relative lowering in vapour pressure...

The relative lowering in vapour pressure is proportional to the ratio of number of

A

Solute molecules to solvent molecules

B

Solvent molecules to solute molecules

C

Solute molecules to the total number of molecules in solution

D

Solvent molecules to the total number of molecules in solution

Text Solution

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The correct Answer is:
C

One mole of electron charge is equivalent to one Faraday.
Given reaction, `CaCl_(2) rarr Ca^(2+) + 2Cl^(-)`
Ca is undergoing oxidation i.e, `Ca^(2+) + 2e^(-) rarr Ca`
We can observe that 40g of Ca takes `2e^(-)` charge, So, 1 mole of Ca(40g)= 2F
Now, 100g of `Ca -= 5F`
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