What is the vapor pressure at 23 degree C of a solution of 1.20g of naphthalene, C10H8 in 25.6 g of benzene, C6H6? The vapor pressure of naphthalene can be neglected. Calculate the vapor-pressure lowering the solution
Do you have the normal vapor pressure of benzene at that temperature? Or have you been told to look it up?
92.40 [mmHg]
delta Psolvent(lowering) = XsolutePosolvent
To calculate the vapor-pressure lowering of a solution, we need to use Raoult's Law. According to Raoult's Law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.
First, we need to calculate the mole fraction of benzene (solvent) in the solution.
The molecular weight of naphthalene (C10H8) is:
C = 12.01 g/mol x 10 = 120.10 g/mol
H = 1.01 g/mol x 8 = 8.08 g/mol
Total molecular weight = 120.10 g/mol + 8.08 g/mol = 128.18 g/mol
The molecular weight of benzene (C6H6) is:
C = 12.01 g/mol x 6 = 72.06 g/mol
H = 1.01 g/mol x 6 = 6.06 g/mol
Total molecular weight = 72.06 g/mol + 6.06 g/mol = 78.12 g/mol
Next, we need to calculate the number of moles for each substance:
Number of moles of naphthalene (C10H8):
Mass of naphthalene = 1.20 g
Number of moles = 1.20 g / 128.18 g/mol = 0.00936 mol
Number of moles of benzene (C6H6):
Mass of benzene = 25.6 g
Number of moles = 25.6 g / 78.12 g/mol = 0.3277 mol
Now, we can calculate the mole fraction of benzene (Xbenzene):
Xbenzene = moles of benzene / total moles
= 0.3277 mol / (0.00936 mol + 0.3277 mol)
= 0.972
Now, we need to find the vapor pressure of pure benzene at 23°C. According to available tables, the vapor pressure of benzene at 23°C is 153.9 mmHg.
Using Raoult's Law, we can calculate the vapor pressure of the solution:
Vapor-pressure lowering = Xbenzene * Vapor pressure of pure benzene
= 0.972 * 153.9 mmHg
≈ 149.6 mmHg
Therefore, the vapor-pressure lowering of the solution is approximately 149.6 mmHg at 23°C.