Question

The nonvolatile, nonelectrolyte urea, CH4N2O (60.1 g/mol), is soluble in water, H2O. How many grams of urea are needed to generate an osmotic pressure of 24.3 atm when dissolved in 216 mL of a water solution at 298 K.

Answer 1

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:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 24.3 atm

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = mass of urea = ? g

$M_{solute}$ = molar mass of urea = 60.1 g/mol

$V_{solution}$ = Volume of solution = 216 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$24.3atm=1\times \frac{m_{solute}\times 1000}{60.1\times 216}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\m_{solute}=\frac{24.3\times 60.1\times 216}{1\times 1000\times 0.0821\times 298}=12.89g$

Hence, the mass of urea needed is 12.89 grams