Question

0,51 g of a compound with a molecular mass of 2,4 x 10^4 a.m.u. is dissolved in waterand made up to 125 ml volume. What will be the osmotic pressure at 25°C?

Answer 1

Answer:

The osmotic pressure is 4.15*10^-3 atm.

Explanation:

1st) The formula of osmotic pressure is:

$\pi=M*R*T$

- π: osmotic pressure.

- : molarity of the solution.

- R: ideal gas constant (0.082 atm.L/mol.K)

- T: temperature on Kelvin scale = 298 K (25°C)

2nd) With the grams andmass of the compound, we can calculate the mnumer of moles

$\begin{gathered} 2.4*10^4g-1mol \\ 0.51g-x=\frac{0.51g*1mol}{2.4*10^4g} \\ x=2.125*10^{-5}mol \end{gathered}$

Now, with the number of moles and volume, we can calculate the molarity of the solution:

$\begin{gathered} 125mL-2.125*10^{-5}mol \\ 1000mL-x=\frac{1000mL*2.125*10^{-5}mol}{125mL} \\ x=1.7*10^{-4} \end{gathered}$

The molarity of the solution is 1.7*10^4M.

3rd) Finally, we have to replace the values in the osmotic pressure formula:

$\begin{gathered} \pi=M*R*T \\ \pi=1.7*10^{-4}\frac{mol}{L}*0.082\frac{atm*L}{mol*K}*298K \\ \pi=4.15*10^{-3}atm \end{gathered}$

So, the osmotic pressure is 4.15*10^3 tm.