Name: Solid Solutions
Uploaded: 2019-07-19
Duration: 2 min 56 s
Description: Solutions can be categorized as solid, liquid, and gaseous based on the physical states of solvents and solutes. In this video, let's learn about solid solutions.

Question

How is the molar mass of a non-volatile substance determined from the lowering of vapour pressure?

Accepted Answer

We know that, for an ideal solution,

$\frac{{p^{o}}_{A} - p_{A}}{{p^{o}}_{A}} = x_{B} x_{B} = \frac{n_{B}}{n_{A} + n_{B}} = \frac{\frac{W_{B}}{M_{B}}}{\frac{W_{A}}{M_{A}} + \frac{W_{B}}{M_{B}}}$

Where n_B = number of moles of solute n_A = number of moles of solvent. W_B and W_A are masses of solute and solvent. M_B and M_A are molecular mass of solute and solvent.

Substituting the value of x_B , we get

$\frac{\frac{W_{B}}{M_{B}}}{\frac{W_{A}}{M_{A}} + \frac{W_{B}}{M_{B}}} = \frac{{p^{o}}_{A} - p_{A}}{{p^{o}}_{A}}$

For the ideal solutions which are dilute, the molar concentration of solute being very small $\frac{W_{B}}{M_{B}} < < \frac{W_{A}}{M_{A}}$ , and can be neglected as compared to that of the solvent. Thus,

$\frac{W_{B}}{M_{B}} + \frac{W_{A}}{M_{A}} ≃ \frac{W_{A}}{M_{A}}$

The equation becomes,

$\frac{\frac{W_{B}}{M_{B}}}{\frac{W_{A}}{M_{A}}} = \frac{{p^{o}}_{A} - p_{A}}{{p^{o}}_{A}} \frac{W_{B} x M_{A}}{W_{A} x M_{B}} = \frac{{p^{o}}_{A} - p_{A}}{{p^{o}}_{A}} M_{B} = \frac{W_{B} x M_{A}}{W_{A}} (\frac{{p^{o}}_{A}}{{p^{o}}_{A} - p_{A}})$

Thus, the molar mass of the solute can be determined if the other quantities in above equation are known.

How is the molar mass of a non-volatile substance determined from the lowering of vapour pressure?

Important Questions on Solutions

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