Question

Helium is collected over water at 25°C and 1.00 atm total pressure. What total volume of gas must be collected to obtain 0.405 g helium? (At 25°C, the vapor pressure of water is 23.8 torr.)

a) 2.50 L

b) 3.00 L

c) 4.00 L

d) 5.00 L

Answer 1

Answer is Option B. To obtain 0.405 g of helium, a total volume of 3.01 L of gas must be collected over water at 25°C and 1.00 atm total pressure.

Step-by-step explanation:

When gases are collected over water, the total pressure is equal to the sum of the partial pressure of the gas and the vapor pressure of water. In this case, the total pressure is 1.00 atm, and the vapor pressure of water at 25°C is 23.8 torr. To find the partial pressure of helium, we subtract the vapor pressure of water from the total pressure: 1.00 atm - 23.8 torr = 0.976 atm. Now we can use the ideal gas law to calculate the volume of helium:

V = (nRT)/P

where V is the volume, n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and P is the pressure. Rearranging the equation to solve for volume, we have:

V = (nRT)/P = (0.405 g / (4.00 g/mol)) * (0.0821 L·atm/(mol·K)) * (298 K) / (0.976 atm) = 3.01 L

Therefore, the total volume of gas that must be collected to obtain 0.405 g of helium is 3.01 L.