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The osmotic pressure of a urea solution ...

The osmotic pressure of a urea solution is 500 mm Hg at `10^@C`. The solution is diluted and its temperature is raised to `25^@C`. It is now found that osmotic pressure of the solution is reduced to 105.3 mm Hg. The extent of dilution of the solution is

A

3 times

B

4 times

C

5 times

D

6 times.

Text Solution

Verified by Experts

The correct Answer is:
C
5 times
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The osmotic presure of a urea solution is 500 mm of Hg at 10^(@)C . The solution is diluted and its temperature is reised to 25^(@)C . It is now found that the osmotic pressure of the solution is reduced to 105.3 mm. Detyermine the extrant of dilution of the solution.

The osmotic pressure of urea solution is 500 mm at 10^(@)C . The solution is diluted and the temperature raised to 25^(@)C , when the osmotic pressure is found to be 105.3 mm . Find the extend if dilution.

Knowledge Check

  • The osmotic pressure of a dilute solution is given by

    A
    `P=P_(o)x`
    B
    `pi V=nRT`
    C
    `Delta P=P_(o)N_(2)`
    D
    `(Delta P)/(P_(o))=(P_(o)-P)/(P_(o))`

  • The osmotic pressure of a dilute solution is given by

    A
    `P=P_(0)xxN_(1)`
    B
    `piV=nRT`
    C
    `DeltaP=P_(0)N_(2)`
    D
    `(DeltaP)/(P^(@))=(P^(@)-P_(s))/P^(@)`

  • Osmotic pressure of urea solution is 56.6 mm Hg at a temperature of 10^(@)C . The solution is diluted and the temperature is raised to 40^(@) C such that the osmotic pressure is found to be 31.3 mm Hg . The extent of dilution will be :

    A
    `V_("final") = 10 V_("initial")`
    B
    `V_("final") = 2 V_("initial")`
    C
    `2V_("final ") = 3V_("initial")`
    D
    `V_("final") = 20 V_("initial")`

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