A solution is prepared by dissolving 0.6 g of urea (molar mass =60g mol^{-1} ) and 1.8 g of glucose (molar mass = 180 g mol^{-1} ) in 100 mL of water at 27^{\circ}C. The osmotic pressure of the solution is:

(R=0.08206 L atm K^{-1}mol^{-1})

  • Option 1)

    8.2 atm

  • Option 2)

    2.46 atm

  • Option 3)

    4.92 atm

  • Option 4)

    1.64 atm

Answers (1)

molar of Urea =\frac{0.6}{60}=\frac{1}{100}                                                     Given Water = 100ml

molar of glucose = \frac{1.8}{180}=\frac{1}{100}                                                  Temp = 300 k

Total moles = \frac{1}{100}+\frac{1}{100}=\frac{2}{100}

Osmotic pressure = \pi =CRT

=\left ( \frac{2}{100} \cdot \frac{1000}{100}\right )\cdot 0.08206\cdot 300

\pi =4.92atm


Option 1)

8.2 atm

Option 2)

2.46 atm

Option 3)

4.92 atm

Option 4)

1.64 atm

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