Question

a) clculate the partial pressure (in atm) of Ne and Ar in the container. gas mixture in a 1.45 L at 298 K container contains 10.0 g of Ne and 10.0 g of Ar. Calculate the partial pressure (in ) of and in the container.

PNe = 4.22 atm , PAr = 4.22 atm

PNe = 8.36 atm , PAr = 4.22 atm

PNe = 8.36 atm , PAr = 8.36 atm

PNe = 4.22 atm , PAr = 8.36 atm

B) What is the molar mass of the gas? A 1.29 g gas sample occupies 688 mL at 37 ∘C and 1.00 atm.

3.87 g/mol

0.258 g/mol

47.7 g/mol

0.0461 g/mol

c) What is the temperature of the gas sample in ∘C at this volume? A gas sample has a volume of 180 mL at 0.00 ∘C. The temperature is raised (at constant pressure) until the volume is 213 mL .

223 ∘C

50 ∘C

323 ∘C

0.00 ∘C

Answer 1

a)

for Neon:

Molar mass of Ne = 20.18 g/mol

mass(Ne)= 10.0 g

use:

number of mol of Ne,

n = mass of Ne/molar mass of Ne

=(10 g)/(20.18 g/mol)

= 0.4955 mol

Given:

V = 1.45 L

n = 0.4955 mol

T = 298.0 K

use:

P * V = n*R*T

P * 1.45 L = 0.4955 mol* 0.08206 atm.L/mol.K * 298 K

P = 8.36 atm

for Ar:

Molar mass of Ar = 39.95 g/mol

mass(Ar)= 10.0 g

use:

number of mol of Ar,

n = mass of Ar/molar mass of Ar

=(10 g)/(39.95 g/mol)

= 0.2503 mol

Given:

V = 1.45 L

n = 0.2503 mol

T = 298.0 K

use:

P * V = n*R*T

P * 1.45 L = 0.2503 mol* 0.08206 atm.L/mol.K * 298 K

P = 4.22 atm

PNe = 8.36 atm , PAr = 4.22 atm

b)

Given:

P = 1.0 atm

V = 688.0 mL

= (688.0/1000) L

= 0.688 L

T = 37.0 oC

= (37.0+273) K

= 310 K

find number of moles using:

P * V = n*R*T

1 atm * 0.688 L = n * 0.08206 atm.L/mol.K * 310 K

n = 2.705*10^-2 mol

mass(solute)= 1.29 g

use:

number of mol = mass / molar mass

2.705*10^-2 mol = (1.29 g)/molar mass

molar mass = 47.7 g/mol

Answer: 47.7 g/mol

c)

Given:

Ti = 0.0 oC

= (0.0+273) K

= 273 K

Vi = 180 mL

Vf = 213 mL

use:

Vi/Ti = Vf/Tf

180.0 mL / 273.0 K = 213.0 mL / Tf

Tf = 323.05 K

Tf = 50.0 oC

Answer: 50.0 oC