The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.198 g sample of liquid CHCl3 is placed in a closed, evacuated 400. mL container at a temperature of 283 K.

Assuming that the temperature remains constant, will all of the liquid evaporate?
What will the pressure in the container be when equilibrium is reached?

The pressure inside the container will be 100 mm Hg when equilibrium is reached.

Use PV = nRT and solve for n, then n = grams/molar mass to solve for grams.
Compare grams in the vapor with grams of the liquid to know if all will have evaporated.

To determine whether all of the liquid chloroform will evaporate, we need to compare the vapor pressure of chloroform at 283 K (given as 100. mm Hg) to the pressure inside the container.

To calculate the number of moles of chloroform in the 0.198 g sample, we need to know the molar mass of chloroform (CHCl3).
The molar mass of CHCl3 is:
1C: 12.01 g/mol
1H: 1.01 g/mol
3Cl: 35.45 g/mol x 3 = 106.35 g/mol

So, molar mass of CHCl3 = 12.01 g/mol + 1.01 g/mol + 106.35 g/mol = 119.37 g/mol.

Now, we can calculate the number of moles of chloroform in the 0.198 g sample using the formula:
moles = mass / molar mass
moles = 0.198 g / 119.37 g/mol = 0.00166 mol

Since the container is closed and initially evacuated, the pressure inside is initially 0 mm Hg.

At equilibrium, the partial pressure of chloroform in the container will be equal to its vapor pressure at 283 K (100. mm Hg).

To calculate the pressure inside the container at equilibrium, we can use the ideal gas law:
PV = nRT

Where:
P = pressure of the gas (unknown)
V = volume of the container (400. mL = 0.4 L)
n = number of moles of the gas (0.00166 mol)
R = ideal gas constant (0.0821 L·atm/(K·mol))
T = temperature in Kelvin (283 K)

Rearranging the equation to solve for P, we have:
P = (nRT) / V
P = (0.00166 mol * 0.0821 L·atm/(K·mol) * 283 K) / 0.4 L
P ≈ 0.24 atm

Converting atm to mm Hg, we have:
P = 0.24 atm * 760 mm Hg/atm = 182 mm Hg

Therefore, when equilibrium is reached, the pressure inside the container will be approximately 182 mm Hg, which is less than the vapor pressure of chloroform (100. mm Hg). This means that not all of the liquid chloroform will evaporate.

To determine if all of the liquid chloroform (CHCl3) will evaporate, we need to compare the vapor pressure of CHCl3 at 283 K to the pressure in the container. If the vapor pressure is higher than the pressure in the container, then all of the liquid will evaporate.

First, let's convert the mass of the CHCl3 sample to moles. The molar mass of CHCl3 is calculated by adding up the atomic masses of its component elements: carbon (C), hydrogen (H), and chlorine (Cl).

Molar mass CHCl3 = (1 mol C) + (1 mol H) + (3 mol Cl) = 12.01 g/mol + 1.01 g/mol + (3 × 35.45 g/mol) = 119.38 g/mol

Next, we can calculate the number of moles of CHCl3 in the sample:

n (moles) = mass (g) / molar mass (g/mol)
n = 0.198 g / 119.38 g/mol = 0.001660 mol

Now, let's calculate the partial pressure of CHCl3 in the container when equilibrium is reached. The partial pressure is the pressure exerted by a single gas component in a mixture.

partial pressure = mole fraction × total pressure

Since the container is evacuated initially, the total pressure is atmospheric pressure (which we'll assume to be 1 atm). The mole fraction can be calculated by dividing the moles of CHCl3 by the total moles of gas in the container.

moles of gas in the container = moles of CHCl3
mole fraction = moles of CHCl3 / total moles of gas = 0.001660 mol / 0.001660 mol = 1

So, the partial pressure of CHCl3 when equilibrium is reached is 1 atm.

Now, let's compare the vapor pressure of CHCl3 at 283 K to the pressure in the container. If the vapor pressure is higher, all of the liquid will evaporate.

Given that the vapor pressure of CHCl3 at 283 K is 100 mm Hg, we need to convert it to atm:

1 atm = 760 mm Hg

vapor pressure (atm) = 100 mm Hg / 760 mm Hg/atm
vapor pressure = 0.1316 atm

Since the vapor pressure (0.1316 atm) is lower than the partial pressure (1 atm), we can conclude that not all of the liquid CHCl3 will evaporate. Some of it will remain in liquid form in the container.

When equilibrium is reached, the pressure in the container will be equal to the partial pressure of CHCl3, which is 1 atm.