INTOR TO CHEMISTRY LLF
INTOR TO CHEMISTRY LLF
5th Edition
ISBN: 9781264501731
Author: BAUER
Publisher: MCG
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Chapter 9, Problem 148QP

(a)

Interpretation Introduction

Interpretation:

The moles and mass of neon gas contained in a 20.0 L container at 389 torr and 300 K are to be calculated.

(a)

Expert Solution
Check Mark

Explanation of Solution

The number of moles of a gas n is equal to the ratio of the amount of gas m present in the container to the molar mass MM of that gas.

n=mMM … (1)

The ideal gas law dictates the behavior of ideal gases. It states that the pressure P and volume V of a gas is directly proportional to the number of moles n present in the gas and the temperature T of the gas.

PV=nRT … (2)

Here, R is the universal gas constant.

Convert 389 torr to atm as shown below:

760 torr=1 atm389 torr=1 atm760 torr×389 torr=0.512 atm

Substitute P as 0.512 atm , T as 300 K , R as 0.08206 L atm mol1 K1 and V as 20.0 L in equation (2) to determine n of neon gas:

0.512 atm×20 L=n×0.08206 L atm mol1 K1×300.0 Kn=0.512 atm×20 L0.08206 L atm mol1 K1×300.0 K=0.416 moles

So, 0.416 mol of neon gas are present. The mass of neon gas is determined by using equation (1). Substitute n as 0.416 mol and MM as 20.18 g mol1 in equation (1):

0.416 mol=m20.18 g mol1m=0.416 mol×20.18 g mol1=8.39 g

(b)

Interpretation Introduction

Interpretation:

The moles and mass of hydrogen gas contained in a 20.0 L container at 389 torr and 300 K are to be calculated.

(b)

Expert Solution
Check Mark

Explanation of Solution

Substitute P as 0.512 atm , T as 300 K , R as 0.08206 L atm mol1 K1 and V as 20.0 L in equation (2) to determine n of hydrogen gas:

0.512 atm×20 L=n×0.08206 L atm mol1 K1×300.0 Kn=0.512 atm×20 L0.08206 L atm mol1 K1×300.0 K=0.416 moles

So, 0.416 mol of hydrogen gas are present. The mass of hydrogen gas is determined by using equation (1). The molecular mass of hydrogen gas is:

MMH2=2×1 g mol1=2 g mol1

Substitute n as 0.416 mol and MM as 2 g mol1 in equation (1) :

0.416 mol=m2 g mol1m=0.416 mol×2 g mol1=0.839 g

(c)

Interpretation Introduction

Interpretation:

The moles and mass of carbon dioxide gas contained in a 20.0 L container at 389 torr and 300 K are to be calculated.

(c)

Expert Solution
Check Mark

Explanation of Solution

Substitute P as 0.512 atm , T as 300 K , R as 0.08206 L atm mol1 K1 and V as 20.0 L in equation (2) to determine n of carbon dioxide gas:

0.512 atm×20 L=n×0.08206 L atm mol1 K1×300.0 Kn=0.512 atm×20 L0.08206 L atm mol1 K1×300.0 K=0.416 moles

So, 0.416 mol of carbon dioxide gas are present. The mass of carbon dioxide gas is determined by using equation (1). The molecular mass of carbon dioxide gas is:

CO2=12 g mol1+216 g mol1=44 g mol1

Substitute n as 0.416 mol , MM as 44 g mol1 in equation (1) :

0.416 mol=m44 g mol1m=0.416 mol×44 g mol1=18.3 g

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Chapter 9 Solutions

INTOR TO CHEMISTRY LLF

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