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**Hint**The Raoult’s law states that the solvent partial vapor pressure in a solution or the mixture is equal to the product of vapor pressure of the pure solvent and mole fraction of the solution. There are either positive deviations for Raoult law and negative deviations for Raoult law. The positive deviation occurs when the vapor pressure of solution is higher than it is expected from Raoult law.

**Complete solution:**

\[{{P}_{total}}={{P}_{A}}+{{P}_{B}}\]……..equation 1

Here \[{{P}_{A}}\]is the vapour pressure of pure solvent A

\[{{P}_{B}}\]Is the vapour pressure of pure solvent B

\[{{P}_{total}}={{P}^{o}}_{A}{{X}_{A}}+{{P}_{B}}^{o}{{X}_{B}}\]……..equation 2

Here \[{{P}^{o}}_{A}\] is the partial pressure of A

\[{{P}_{B}}^{o}\]Is the partial pressure of B

\[{{X}_{A}}\]Is the mole fraction of A

\[{{X}_{B}}\]Is the mole fraction of B

The equation given to us is

\[{{P}_{total}}=220-110{{X}_{B}}\]…….equation3

On comparing equation 3 with equation 1 we get

\[{{P}_{A}}+{{P}_{B}}\]=\[220-110{{X}_{B}}\]

So from this we came to know that \[{{P}_{A}}\]=220 and\[{{P}_{B}}=-110{{X}_{B}}\] …..equation 4

On simplifying equation 2 we get one more formula and that is

\[{{P}_{total}}={{X}_{B}}({{P}_{B}}^{o}-{{P}^{o}}_{A})+{{P}^{o}}_{A}\]

Now on simplifying we get

\[\begin{align}

& {{X}_{B}}({{P}_{B}}^{o}-{{P}^{o}}_{A})=-110{{X}_{B}} \\

& \Rightarrow {{P}_{B}}^{o}=-110+220 \\

& \Rightarrow {{P}_{B}}^{o}=110 \\

& \\

\end{align}\]

**Hence, the correct option is (B).**

**Note:**Raoult law is widely used in estimating the amount of the individual component in the solution. It is applicable to the solutions which are made up by non volatile solutes. Raoult law is not applicable to those solutes which get dissociated or associated in the solution.

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