Help with this chemistry question. Thanks in advance.

Using equation:

Help with this chemistry investigation. Thanks in space.

The exhalation of acetic hard contains individual and embrace molecules in makeweight. (CH_3COOH)_2 = 2CH_3COOH At 25 extent C and 0.020 atm hurry, the PV emanation coercion 60 g of acetic hard exhalation is 0.541RT, and 40 extent C and 0.020 atm, is 0.593RT. Calculate the faction of the exhalation coercionming individual molecules at each atmosphere and the esteem coercion the makeweight faithful at atmosphere. K_P = P^2_CH_3COOH/P_(CH_3COOH)_2

Using equation:

PV = nRT

At 25^{0}C, we accept:

PV = 0.541RT

So,

n = 0.541

Assume that at this temp, we accept ‘x’ moles of individual molecules and ‘y’ moles of embrace molecules. So we accept the aftercited span equations:

x + y = 0.541

60*x + 120*y = 60

Solving these equations we get:

x = 0.082, y = 0.459

So, mole faction of individual molecules at this temp = 0.082/0.541 = **0.151**

Makeweight faithful at this temp = (0.151*0.020)^{2}/(0.849*0.020) = **0.000537 atm**

At 40^{0}C, we accept:

PV = 0.593RT

So,

n = 0.593

Assume that at this temp, we accept ‘x’ moles of individual molecules and ‘y’ moles of embrace molecules. So we accept the aftercited span equations:

x + y = 0.593

60*x + 120*y = 60

Solving these equations we get:

x = 0.186, y = 0.407

So, mole faction of individual molecules at this temp = 0.186/0.593 = **0.314**.

Makeweight faithful at this temp = (0.314*0.020)^{2}/(0.686*0.020) = **0.00287 atm**