At 40 degreee celcius the vapour pressure, in torr, of methyl aclohol-ethyl alcohol solutions is represented by P= 119x + 135 where x is the mole fraction of methyl alcohol. what are vapours pressures of the pure components at this tempreature?

Let, vapour pressure of methyl alcohol is p_m ethyl alcohol has p_​e . Similarly, the mole fractions are x_m and x_e then we have,
final pressure of solution :
$$\begin{gathered} P=p_m^0{x}_m +p_e^0{x}_e =p_m^0{x}_m + p_e^0(1-x_m) \\ Or, P=(p_m^0-p_e^0) x_m + p_e^0 \end{gathered}$$
Now, we have, the pressure of methyl alcohol is , =119x+135, then equating the value in avbove equation we get:
$$\begin{gathered} P=(p_m^0-p_e^0) x_m + p_e^0 \\ Or, P =119x+ 135 \\ So, that, p_e^0 = 135 torr \\ (p_m^0-p_e^0) = 119 \\ So, p_m^0 = 119 +135 = 245 torr \end{gathered}$$
Hence, the pure vapour pressure of ethyl alcohol is 135torr and pure methyl alcohol is 245 torr.