Question

Derive the formula to determine the molar mass of solute on

the basis of Raoult’s law for ideal solution prepared by the

dissolving non-volatile solute in volatile solvent.​

Answer 1

Answer:

Raoult's law (/ˈrɑːuːlz/ law) is a law of thermodynamics established by French chemist François-Marie Raoult in 1887.[1] It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture. In consequence, the relative lowering of vapour pressure of a dilute solution of nonvolatile solute is equal to the mole fraction of solute in the solution.

Mathematically, Raoult's law for a single component in an ideal solution is stated as

{\displaystyle p_{i}=p_{i}^{\star }x_{i},} {\displaystyle p_{i}=p_{i}^{\star }x_{i},}

where {\displaystyle p_{i}} p_{i} is the partial pressure of the component {\displaystyle i} i in the gaseous mixture (above the solution), {\displaystyle p_{i}^{\star }} {\displaystyle p_{i}^{\star }} is the equilibrium vapor pressure of the pure component {\displaystyle i} i, and {\displaystyle x_{i}} x_{i} is the mole fraction of the component {\displaystyle i} i in the mixture (in the solution).[2]

Where two volatile liquids A and B are mixed with each other to form a solution, the vapour phase consists of both components of the solution. Once the components in the solution have reached equilibrium, the total vapor pressure of the solution can be determined by combining Raoult's law with Dalton's law of partial pressures to give

{\displaystyle p=p_{\text{A}}^{\star }x_{\text{A}}+p_{\text{B}}^{\star }x_{\text{B}}+\cdots .} {\displaystyle p=p_{\text{A}}^{\star }x_{\text{A}}+p_{\text{B}}^{\star }x_{\text{B}}+\cdots .}

If a non-volatile solute (zero vapor pressure, does not evaporate) is dissolved into a solvent to form an ideal solution, the vapor pressure of the final solution will be lower than that of the solvent. The decrease in vapor pressure is directly proportional to the mole fraction of solute in an ideal solution;

{\displaystyle p=p_{\text{A}}^{\star }x_{\text{A}},} {\displaystyle p=p_{\text{A}}^{\star }x_{\text{A}},}

{\displaystyle \Delta p=p_{\text{A}}^{\star }-p=p_{\text{A}}^{\star }(1-x_{\text{A}})=p_{\text{A}}^{\star }x_{\text{B}}.} {\displaystyle \Delta p=p_{\text{A}}^{\star }-p=p_{\text{A}}^{\star }(1-x_{\text{A}})=p_{\text{A}}^{\star }x_{\text{B}}.}

Answer 2

To derive:

Formula to determine the molar mass of solute on the basis of Raoult's Law for ideal solution prepared by dissolving non volatile solute in volatile solvent.

Derivation:

Let P°_(A) be vapour pressure of pure solvent and P_(A) be vapour pressure of solution.

B represents Solute and A represents Solvent.

We know that :

$\therefore \: \dfrac{P_{A} \degree - P_{A}}{P_{A} \degree} = x_{B}$

$= > \: \dfrac{P_{A} \degree - P_{A}}{P_{A} \degree} = \dfrac{ (\frac{W_{B}}{M_{B}}) }{ \frac{W_{B}}{M_{B}} + \frac{W_{A}}{M_{A}} }$

For a dilute solution , we can say that:

$\boxed{n_{A} \: > > \: n_{B}}$

So, continuing with the Equation:

$= > \: \dfrac{P_{A} \degree - P_{A}}{P_{A} \degree} = \dfrac{ (\frac{W_{B}}{M_{B}}) }{ ( \frac{W_{A}}{M_{A}} ) }$

$= > \: \dfrac{P_{A} \degree - P_{A}}{P_{A} \degree} = (\dfrac{W_{B}}{M_{B}}) \times (\dfrac{M_{A}}{W_{A}} )$

$= > \: \dfrac{P_{A} \degree - P_{A}}{P_{A} \degree} = (\dfrac{1}{M_{B}}) \times (\dfrac{M_{A} \times W_{B}}{W_{A}} )$

$= > \: M_{B} = \bigg \{\dfrac{P_{A} \degree}{P_{A} \degree - P_{A}} \bigg \} (\dfrac{M_{A} \times W_{B}}{W_{A}} )$

So, final answer is :

$\boxed{ \red{ \bold{\: M_{B} = \bigg \{\dfrac{P_{A} \degree}{P_{A} \degree - P_{A}} \bigg \} (\dfrac{M_{A} \times W_{B}}{W_{A}} )}}}$