Chapter 28, Problem 28.11P

(a)

Interpretation Introduction

Interpretation:

The mass of a single particle of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and the number of particles if ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture similarly for ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ has to be determined.

Explanation of Solution

The mass of single particle of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ can be determined by knowing its volume and then from that volume we can find the mass as follows.

$Volume=\frac{4}{3}\pi r^3$

Where, $r=0.075mm=7.5\times {10}^{-3}cm.$

Hence, $Volume=1.767\times {10}^{-6}mL.$

$\begin{array}{l}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{mass}=\left(1.767\times {10}^{-6}mL\right)\left(2.532\times g/mL\right)=4.474\times {10}^{-6}g.\\ {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\text{mass}=\left(1.767\times {10}^{-6}mL\right)\left(2.428\times g/mL\right)=4.291\times {10}^{-6}g\end{array}$

The number of particles of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}=\left(4.00g\right)/\left(4.474\times {10}^{-6}\text{g/particle}\right)=8.941\times {10}^5.$

The number of particles of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}=\left(96.00g\right)/\left(4.291\times {10}^{-6}\text{g/particle}\right)=2.237\times {10}^7.$

The fraction of each type is calculated as follows.

$p{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}=\left(8.941\times {10}^5\right)/\left(8.941\times {10}^5+2.237\times {10}^7\right)=0.0384$

$q{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}=\left(2.237\times {10}^7\right)/\left(8.941\times {10}^5+2.237\times {10}^7\right)=\mathrm{0.962.}$

(b)

Interpretation Introduction

Interpretation:

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

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 $n=2.326\times {10}^4.$

Therefore, the expected number of particles in the given mixture is $n=2.326\times {10}^4.$

(c)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

The expected number of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ particles in 0.100g is 1/1000 of number in 100grams = $8.94\times {10}^2.$

The expected number of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ particles in 0.100g is 1/1000 of number in 100grams = $2.24\times {10}^4.$

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

$\text{Sampling}\text{standard}\text{deviation = }\sqrt{npq}=\sqrt{(2.326\times {10}^4)(0.0384)(0.962)}=29.3$

Now,

$\text{Relative}\text{sampling}\text{standard}\text{deviation}\text{for}{\text{Na}}_{\text{2}}\text{CO}{\text{3}}_{}=\frac{29.3}{8.94\times {10}^2}=3.28\%.$

$\text{Relative}\text{sampling}\text{standard}\text{deviation}\text{for}{\text{K}}_{\text{2}}\text{CO}{\text{3}}_{}\text{=}\frac{\text{29}\text{.3}}{\text{2}{\text{.24×10}}^4}\text{=0}\text{.131\%}\text{.}$