Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
Question
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Chapter 28, Problem 28.11P

(a)

Interpretation Introduction

Interpretation:

The mass of a single particle of Na2CO3 and the number of particles if Na2CO3 in the mixture similarly for K2CO3 has to be determined.

(a)

Expert Solution
Check Mark

Explanation of Solution

The mass of single particle of Na2CO3 and K2CO3 can be determined by knowing its volume and then from that volume we can find the mass as follows.

Volume=43πr3

Where, r=0.075mm=7.5×103cm.

Hence, Volume=1.767×106mL.

Na2CO3mass=(1.767×106mL)(2.532×g/mL)=4.474×106g.K2CO3mass=(1.767×106mL)(2.428×g/mL)=4.291×106g

The number of particles of Na2CO3=(4.00g)/(4.474×106g/particle)=8.941×105.

The number of particles of K2CO3=(96.00g)/(4.291×106g/particle)=2.237×107.

The fraction of each type is calculated as follows.

pNa2CO3=(8.941×105)/(8.941×105+2.237×107)=0.0384

qK2CO3=(2.237×107)/(8.941×105+2.237×107)=0.962.

(b)

Interpretation Introduction

Interpretation:

The expected number of particles in 0.100g of the given mixture has to be determined.

(b)

Expert Solution
Check Mark

Explanation of Solution

The expected number of particles in 0.100g of the given mixture can be calculated as follows.

The given number of particles is 0.100 and is n=2.326×104.

Therefore, the expected number of particles in the given mixture is n=2.326×104.

(c)

Interpretation Introduction

Interpretation:

The relative sampling standard deviation for the given information has to be determined.

(c)

Expert Solution
Check Mark

Explanation of Solution

Before going to answering let’s find out the following values.

The expected number of Na2CO3 particles in 0.100g is 1/1000 of number in 100grams = 8.94×102.

The expected number of K2CO3 particles in 0.100g is 1/1000 of number in 100grams = 2.24×104.

Now, we have to determine the sampling standard deviation as follows.

Samplingstandarddeviation = npq=(2.326×104)(0.0384)(0.962)=29.3

Now,

RelativesamplingstandarddeviationforNa2CO3=29.38.94×102=3.28%.

RelativesamplingstandarddeviationforK2CO3=29.32.24×104=0.131%.

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