   Chapter 13, Problem 93QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
1 views

# What volume does a mixture of 14 . 2 g of He and 21 . 6 g of H 2 Occupy at 28  ° C and 0. 985 atm?

Interpretation Introduction

Interpretation:

The volume of mixture of He and H2 gas should be calculated.

Concept Introduction:

According to ideal gas equation:

PV=nRT

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

The value of Universal gas constant can be taken as 0.082 L atm K1 mol1.

Mass of gas can be calculated from number of moles and molar mass as follows:

m=n×M

Here, n is number of moles, m is mass and M is molar mass.

Explanation

Given Information:

The mass of He gas is 14.2 g, mass of H2 gas is 21.6 g, temperature is 28 C and pressure is 0.985 atm.

Calculation:

From given mass of He and hydrogen gas, first calculate the number of moles as follows:

n=mM

Molar mass of He is 4 g/mol thus,

nHe=14.2 g4 g/mol=3.55 mol

Molar mass of H2 gas is 2 g/mol thus,

nH2=21.6 g2 g/mol=10.8 mol

Total number of moles of He and hydrogen gas will be:

nT=nHe+nH2

Putting the values,

nT=3.55 mol+10

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