Question

The osmotic pressure of solution obtained by mixing 200cm^3 of 2%(mass-volume)solution of urea with 200cm^3 of 3.42%solution of sucrose at 20°c is

Answer 1

The osmotic pressure of a solution is 2.586 atm

Explanation:

We are given:

2% m/v urea solution

This means that 2 g of urea is present in 100 ml of solution

3.42% m/v sucrose solution

This means that 3.42g of sucrose is present in 100 ml of solution

The number of moles is calculated by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ...(1)

Molar mass of urea = 60 g/mol

Molar mass of sucrose = 342 g/mol

Using equation 1:

$\text{Moles of urea}=\frac{2g}{60g/mol}=0.033mol$

$\text{Moles of sucrose}=\frac{3.42g}{342g/mol}=0.01mol$

Total number of moles = [0.033 + 0.01] = 0.043 moles

Osmotic pressure is calculated by using the equation:

$\pi=MRT$

OR

$\pi=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in }cm^3}\times RT$          ...(2)

where,

$\pi$ = osmotic pressure

R = Gas constant = 0.0821 L.am/mol.K

T = temperature = $20^oC=[20+273]= 293K$

Volume of solution = $[200+200]cm^3=400cm^3$

Moles of solute = 0.043 moles

Putting values in equation 2, we get:

$\pi=\frac{0.043\times 1000}{400}\times 0.0821\times 293\\\\\pi=2.586atm$

Hence, the osmotic pressure of solution is 2.586 atm

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