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What volume will 15.4 grams of nitrogen gas, N2, occupy at 2.02 atm and 15 °C? The molar mass of nitrogen gas is 28.01348 g/mole.

A.373 L
B.746 L
C.7.46 L
D.6.43 L

User Fstopzero
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2 Answers

3 votes

Answer:

The correct answer is D. 6, 43L.

Step-by-step explanation:

First we calculate the number of moles in 14 grams of Nitrogen using the molar mass, we convert the unit of temperature in Celsius into Kelvin and then we apply the ideal gas law by solving for volume (using the ideal gas constant R = 0.082 l atm / K mol):

28,01348 g----1 mol N2

15,4g ----x= (15,4g x1 mol N2)/28,01348 g= 0,55 mol N2

0°C= 273K ----> 15°C= 273 + 15= 288 K

PV= nRT ---> V= (nRT)/P

V= (0,55 mol x 0,082 l atm/ K molx 288K)/2,02 atm

V=6, 43 L

User Twinkle
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4.8k points
0 votes

Answer: Option D) 6.43 L

Step-by-step explanation:

Given that,

volume (V) = ?

Pressure (P) = 2.02 atm

Temperature (T) = 15 °C

[Convert 15°C to Kelvin by adding 273

15°C + 273 = 288K]

Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1

Number of moles of N2 (n) = ?

molar mass of N2 (m.m) = 28.01348 g/mole

mass in grams of N2 = 15.4 grams

Recall that Number of moles

= mass in grams / molar mass

n = 15.4 grams / 28.01348 g/mole

n = 0.55 moles

Then, apply ideal gas equation

pV = nRT

2.02 atm x V = 0.55 moles x (0.0821 atm L K-1 mol-1 x 288K)

2.02 atm•V = 13 atm•L

Divide both sides by 2.02 atm

2.02 atm•V/2.02 atm = 13 atm•L/2.02 atm

V = 6.43 L

Thus, the volume of nitrogen gas is 6.43 liters

User Sujoy Gupta
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