Liquids A and B form an ideal solution at 30°C, the total vapour pressure of a solution containing 1 mol of A and 2 mol of B is 250 mmHg. The total vapour pressure becomes 300 mmHg when 1 more mol of A is added to the first solution. The vapour pressures of pure A and B at the same temperature are
150, 450 mmHg
125, 150 mmHg
450, 150 mmHg
250, 300 mmHg
Let vapour pressure of \(A = P^0_A\) and vapour pressure of \(B = P^0_b\)
In the first case,
Mole fraction of \(A\left(x_{A}\right)=\frac{1}{1+2}=\frac{1}{3}\)
Mole fraction of \(B\left(x_{B}\right)=\frac{2}{1+2}=\frac{2}{3}\)
According to Raoult's law,
Total vapour pressure \(=250=P^{0}_{A}x_{A}+P^{0}_{B}x_{B}\)
\(250=\frac{1}{3}P^{0}_{A}+\frac{2}{3}P^{0}_{B} \,...\left(i\right)\)
In the second case,
Mole fraction of \(A\left(x_{A}\right)=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}\)
Mole fraction of \(B\left(x_{B}\right)=\frac{2}{4}=\frac{1}{2}\)
\(\therefore\) Total vapour pressure
\(300=P^{0}_{A}x_{A}+P^{0}_{B}x_{B} \)
\(300=\frac{1}{2}P^{0}_{A}+\frac{1}{2}P^{0}_{B}...(ii)\)
Multiplying equation \((i)\) by \(\frac{1}{2}\) and equation \(\left(ii\right)\) by \(\frac{1}{3}\)
\(\frac{1}{6}P^{0}_{A}+\frac{2}{6}P^{0}_{A}=125\)
\(\frac{\frac{1}{6}P^{0}_{A}+\frac{1}{6}P^{0}_{B}=100}{\frac{1}{6}P^{0}_{B}=25}\)
\(P^{0}_{B}=25\times6=150\,mm \,Hg\)
Now, substitute the value of \(P^{0}_{B}=25\) in equation (ii), we get
\(300=P^{0}_{A}\times\frac{1}{2}+150\times\frac{1}{2}\)
\(P^{0}_{A}=450\,mmHg\)
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
The solutions can be classified into three types:
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types: