Question

A gaseous product of a reaction is collected at 280K and 0.95 atm. Given R=0.0821L⋅atmmol⋅K, what is the molar mass of the gas, in grams per mole, if 5.49 g of gas occupies 4.92 L?
A. 0.04 g/mol

B. 133 g/mol

C. 24 g/mol

D. 27 g/mol

Answer 1

Answer:

I took the test, see below

Explanation:

1. The volume increases to twice its original value.

2. volume & temperature are directly proportional

volume & pressure are inversely proportional

pressure & temperature are directly proportional

3. Gay-Lussac’s law, by seeing how changes in temperature affect the pressure of the gas

4.V1P1/T1=V2P2/T2

5. It is a straight line with a positive slope showing that an increase in temperature results in an increase in volume.

6. keeping the pressure constant and increasing the temperature

7. When temperature is held constant and volume increases, the pressure increases.

8. pressure, volume, temperature, number of moles

9. volume

10. 7.10 L/mol

11. 0.105 mol

12. 27 g/mol

13. the temperature increasing by a factor of 2

The rest are short answers, just give those your best shot.

Answer 2

The molar mass of the gas in grams per mole is 27 g/mol

The correct answer to the question is Option D. 27 g/mol

To solve this question, we'll begin by calculating the number of mole of the gas. This can be obtained as follow:

Temperature (T) = 280 K

Pressure (P) = 0.95 atm

Volume (V) = 4.92 L

Gas constant (R) = 0.0821 L⋅atm/mol⋅K

Number of mole (n) =?

PV = nRT

0.95 × 4.92 = n × 0.0821 × 280

4.674 = n × 22.988

Divide both side by 22.988

n = 4.674 / 22.988

n = 0.203 mole

Thus, the number of mole of the gas is 0.203 mole.

Finally, we shall determine the molar mass of the gas. This can be obtained as follow:

Number of mole of gas = 0.203 mole.

Mass of gas = 5.49 g

Molar mass =?

Mole = mass / molar mass

0.203 = 5.49 / molar mass

Cross multiply

0.203 × molar mass = 5.49

Divide both side by 0.203

Molar mass = 5.49 / 0.203

Molar mass = 27 g/mol

Therefore, the molar mass of the gas is 27 g/mol

Option D. 27 g/mol gives the correct answer to the question.

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