In an experiment, a student collects 107 mL of H2 over water at a pressure of 104.8 kPa and a temperature of 30°C. How many grams of H2 are present?

Question

A) 0.024 g

B) 4.00 g

C) 0.04 g

D) 0 049 g

And why

Answer 1

As per @Anthony T., I went ahead and did the problem using different values for R and after converting pressure to atmospheres. I got the same answer as he did, which does not match any of the choices provided.

PV = nRT and solve for n (moles) of H₂

P = 104.8 kPa x 1 atm/101.325 kPa = 1.034 atm

V = volume = 107 ml = 0.107 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temp in K = 30 + 273 = 303K

Before using the pressure of 1.034 atm, we must subtract the vapor pressure of water at 30ºC which I found to be 0.0418 atm (looked it up). So, P = 1.034 atm - 0.0418 atm = 0.9922 atm

PV = nRT

n = PV/RT = (0.9922 atm)(0.107 L) / (0.0821 Latm/Kmol)(303K)

n = 0.00427 moles

mass = 0.00427 moles x 2 g/mol = 0.00854 g = 0.009 g

Answer 2

I worked out the problem but didn't get any of the answers you listed. Here is what I did.

As we know the volume, temperature, and pressure of the gas, we can use the ideal gas equation which is

PV = nRT where P is pressure, V is volume, n is number of moles, T is the absolute temperature, and R is the appropriate value of the gas constant.

As the pressure is in kPa and volume is in mL, I looked up the value to be used for R which turned out to be 8314.5 cm³ kPa/mole-deg. Also, the vapor pressure of water contributes a small amount of pressure to the gas; so, it should be subtracted form the total pressure. The vapor pressure of water in kPa at 30 C was found to be 0.557 kPa. Subtracting this from 104.8 kPa gives 104.2 kPa.

Plugging this into the ideal gas equation and solving for n yields PV/RT or 104.2 x 107 /(8314.5 x 303)

which calculates to 0.00443 moles. Converting to grams by 2.016 g/mole x 0.00443 moles gives 0.009 g of H₂ which doesn't match any of your choices.

I would like some other tutors to try this problem.