The problem states that 113 L of helium gas at 1360C and prevailing barometric pressure is passed through molten silver at the same temperature. The gas becomes saturated with silver vapor, and as a result, the liquid silver loses 0.120 g in mass. We need to determine the vapor pressure of liquid silver at 1360C.

To solve this problem, we can use the ideal gas law and the concept of vapor pressure. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Let's break down the problem into steps:

Step 1: Convert the temperature from Celsius to Kelvin.
1360C + 273 = 1633 K

Step 2: Calculate the number of moles of helium gas.
We can use the ideal gas law to determine the number of moles of helium gas. Since the gas is at the same temperature and pressure before and after passing through the silver, the number of moles remains constant.

PV = nRT
n = PV/RT

Given:
P = the prevailing barometric pressure
V = 113 L
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = 1633 K

Calculate n using the given values.

Step 3: Calculate the change in mass of the liquid silver.
The problem states that the liquid silver loses 0.120 g in mass. This loss in mass is due to the vaporization of silver. We can use the molar mass of silver to calculate the number of moles of silver vaporized.

The molar mass of silver (Ag) is 107.87 g/mol.

Step 4: Calculate the vapor pressure of liquid silver.
The vapor pressure of a substance is the pressure exerted by its vapor when it is in equilibrium with its liquid or solid phase. In this case, the helium gas becomes saturated with silver vapor, indicating that the vapor pressure of silver is equal to the partial pressure of helium.

Since the number of moles of helium gas remains constant, we can use the ideal gas law to determine the partial pressure of helium.

PV = nRT

Given:
P = the prevailing barometric pressure
V = 113 L
n = number of moles of helium gas (calculated in Step 2)
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = 1633 K

Calculate P using the given values.

Step 5: Convert the pressure to mm Hg.
The vapor pressure is typically reported in mm Hg. To convert from atm to mm Hg, we can use the following conversion factor:

1 atm = 760 mm Hg

Multiply the pressure obtained in Step 4 by 760 to convert it to mm Hg.

Step 6: Determine the correct answer.
Compare the obtained vapor pressure with the given answer. If the obtained vapor pressure matches the given answer, then the answer is correct.