Answer: The mass of urea needed is 12.89 grams
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
or,
[tex]\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 24.3 atm
i = Van't hoff factor = 1 (for non-electrolytes)
[tex]m_{solute}[/tex] = mass of urea = ? g
[tex]M_{solute}[/tex] = molar mass of urea = 60.1 g/mol
[tex]V_{solution}[/tex] = Volume of solution = 216 mL
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = 298 K
Putting values in above equation, we get:
[tex]24.3atm=1\times \frac{m_{solute}\times 1000}{60.1\times 216}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\m_{solute}=\frac{24.3\times 60.1\times 216}{1\times 1000\times 0.0821\times 298}=12.89g[/tex]
Hence, the mass of urea needed is 12.89 grams
