Question

A 1.00 L flask is filled with 1.10 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.100 atm .1. Part AWhat is the partial pressure of argon, PAr, in the flask?Express your answer to three significant figures and include the appropriate units.2. Part BWhat is the partial pressure of ethane, Pethane, in the flask?Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer: a) Partial pressure of argon is 0.673 atm.

b) Partial pressure of ethane is 0.427 atm.

Explanation:

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the argon gas = ?

V= Volume of the gas = 1.00 L

T= Temperature of the gas = 25°C = 298 K       (0°C = 273 K)

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{1.10g}{40g/mol}=0.0275moles$

$P=\frac{nRT}{V}=\frac{0.0275\times 0.0821\times 298}{1.00}=0.673atm$

Thus the partial pressure of argon is 0.673 atm.

b) According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$

$p_{total}=p_{Ar}+p_{ethane}$

$1.100=0.673+p_{ethane}$

$1.100-0.673=p_{ethane}$

$p_{ethane}=0.427atm$

Thus partial pressure of ethane is 0.427 atm.