Two liquids X and Y form an ideal solution. At 300 K, the vapor pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, the vapor pressure of the solution increases by 10 mmHg. What are the vapor pressures (in mmHg) of X and Y in their pure states, respectively?

Let P0X and P0Y be the vapor pressures of X and Y in their pure states, respectively.

Initially, the solution contains 1 mol of X and 3 mol of Y. The mole fractions are:

Mole fraction of X (χX) = 1 mol / (1 mol + 3 mol) = 0.25
Mole fraction of Y (χY) = 3 mol / (1 mol + 3 mol) = 0.75

According to Raoult's law for an ideal solution:

Ptotal = χX * P0X + χY * P0Y

550 mmHg = 0.25 * P0X + 0.75 * P0Y (Equation 1)

After adding 1 mol of Y, the solution contains 1 mol of X and 4 mol of Y. The new mole fractions are:

Mole fraction of X (χX) = 1 mol / (1 mol + 4 mol) = 0.2
Mole fraction of Y (χY) = 4 mol / (1 mol + 4 mol) = 0.8

The new total vapor pressure is 550 mmHg + 10 mmHg = 560 mmHg.

560 mmHg = 0.2 * P0X + 0.8 * P0Y (Equation 2)

Now we have a system of two equations with two unknowns (P0X and P0Y):

0.25 * P0X + 0.75 * P0Y = 550
0.2 * P0X + 0.8 * P0Y = 560

We can solve this system of equations using substitution or elimination. Let's use elimination:

Multiply Equation 1 by 8 and Equation 2 by -7.5:

2 * P0X + 6 * P0Y = 4400
-1.5 * P0X - 6 * P0Y = -4200

Add the two equations:

0.5 * P0X = 200
P0X = 400 mmHg

Substitute P0X = 400 mmHg into Equation 1:

0.25 * 400 + 0.75 * P0Y = 550
100 + 0.75 * P0Y = 550
0.75 * P0Y = 450
P0Y = 600 mmHg

Therefore, the vapor pressures of X and Y in their pure states are 400 mmHg and 600 mmHg, respectively.