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**Tip:**Follow these simple tips to find the total vapour pressure over the solution phase from the given data:

Say you are given two liquids and both are volatile. When we mix the two liquids, both the liquids will eventually evaporate and exert a pressure over the solution phase.

To make things easier, we are already given the values of the vapour pressure of the pure components. Plus, we are given the amounts of the two components which have been mixed to form an ideal solution.

**Step 1-**First of all, we will find the number of moles of each component

For this, we will find the molar mass of each component, and then we can find the number of moles of each component by applying this expression

$n=MW $

Here, W is the Weight of the component, and M is Molar mass of the component

**Step 2**- Once we get the values of number of moles of component 1 and component 2, we can find the mole fraction of component 1 and component 2

$x_{A}=n_{A}+n_{B}n_{A} $

$x_{B}=1−x_{A}$

**Step 3**-Now, go on and apply Raoult's law to find the partial vapour pressure of each component:

$p_{A}=x_{A}p_{A}$

**(Equation 1)**$p_{B}=x_{B}p_{B}$

**(Equation 2)****Step 4**-Finally, add equations 1 and 2 to find the total pressure:

$p_{total}=p_{A}+p_{B}=x_{A}p_{A}+x_{B}p_{B}$

Now, let's practice this question to become a pro!

Vapour pressure of chloroform $(CHCl_{3})$ and dichloromethane $(CH_{2}Cl_{2})$ at $25_{o}C$ are 200 mm Hg and 415 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of $CHCl_{3}$ and 40 g of $CH_{2}Cl_{2}$ at the same temperature will be:

[Molecular mass of $CHCl_{3}=119.5g/mol$ and molecular mass of $CH_{2}Cl_{2}=85g/mol$]

[Molecular mass of $CHCl_{3}=119.5g/mol$ and molecular mass of $CH_{2}Cl_{2}=85g/mol$]