Question

A) A 1.00 L flask is filled with 1.25 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.450 atm .
What is the partial pressure of argon, PAr, in the flask?
What is the partial pressure of ethane, Pethane, in the flask?

b)To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.6-L bulb, then filled it with the gas at 1.30 atm and 20.0  ∘C and weighed it again. The difference in mass was 3.9 g . Identify the gas.

Answer 1

A) ideal gas lawequation P = nRT / V

n(Ar) = (1.25 g Ar) / (39.948 g Ar/mol)

R = 0.08205746 L atm/K mol

T = 273+25 = 298 K

V = 1 L

so P = (1.25) / (39.948) x (0.08205746) x (298) K / (1.00 ) = 0.765 atm

(I assume pentane is the mistake)

P(total) = P(Ar) + P(ethane)

1.450 = 0.765 + P(ethane) =====> P(ethane) = 0.685 atm

B)

PV=nRT

P =  1.30 atm, V = 2.6 L, n = ? , R = 0.0821 L atm/ (mol K), T = 20+273 = 293 K

so n = PV/RT ; n = (1.30*2.6) / (0.0821*293) = 0.14 moles

we know that n = wt/ mol wt from this we can find out mol wt of the diatomic compound. (wt was given 3.9 g)

0.14 = 3.9 / mol wt ===> mol wt = 28

N2 mol wt also 28 from piriodic table that means, the filled gas is N2.