Question

At $27^{\circ} C$, a solution containing $25 \,g$ of solute in $2500 \,mL$ of solution exerts an osmotic pressure of $400 \,Pa$. The molar mass of the solute is ____ $g\,mol ^{-1}$. (Nearest integer)
(Given : $R =0.083 \,L\,bar\, K ^{-1} \,mol ^{-1}$ )

Answer 1

Correct Answer: 62250

\(\pi=CRT \)
\(\frac{400Pa}{10^5}​=\frac{\frac{​2.5g}{M_o}​}{\frac{250}{1000L}}​\times0.83\frac{L−bar​}{K⋅mol}\times300K \)
\(M_o​=62250\)

So, the answer is 62250.

Answer 2

The osmotic pressure (\(\pi\)) of a solution is given by the formula:
\(\pi = MRT\)
where M is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin.

To find the molar mass of the solute, we need to first find the molarity of the solution:
\(\text{Molarity }(M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)

We are given that the solution contains 2.5g of solute in 250.0mL of solution. To convert mL to L, we divide by 1000:
\(\text{Volume of solution} = \frac{250.0 mL}{1000} = 0.250 L\)

To find the moles of solute, we need to use the formula:
\(\text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass}}\)

Let's assume that the molar mass of the solute is x g/mol. Then we have:
\(\text{moles of solute} = \frac{2.5 g}{x} g/mol\)

Now we can calculate the molarity of the solution:
\(\text{Molarity }(M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)
\(M = \frac{\frac{2.5 g}{x g/mol}}{0.250 L}\)
\(M = \frac{10}{x} mol/L\)

Next, we can use the given osmotic pressure and temperature to find the molarity of the solution:
\(\pi = MRT\)
\(400 Pa = \frac{10}{x} mol/L \times 0.083 L bar K^{-1} mol^{-1} \times (27°C + 273.15) K\)
Simplifying and solving for x, we get:
\(x = 165\ gmol^{-1}\) (nearest integer)

Therefore, the molar mass of the solute is \(165 \ gmol^{-1}\), i.e., 62250.