Chapter 2, Problem 101A

Interpretation Introduction

Interpretation:

From the given data, the density of the sample needs to be determined. If the accepted density of the sample is $6.95\text{g/mL}$, percent error in the calculation needs to be determined.

Concept introduction:

Density of a sample of matter or an object is equal to the mass of the sample divided by the volume of the sample.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

Answer to Problem 101A

The density and the percent error of the sample is 6.82 g/mL and 1.87% respectively.

Explanation of Solution

Given Information:

The given data is as follows:

Volume and Mass Data
Mass of sample	$20.46\text{g}$
Volume of water	$40.0\text{mL}$
Volume of water + sample	$43.0\text{mL}$

Calculation:

Mass of the sample = $20.46\text{g}$

Volume of the water = $40.0\text{mL}$

Volume of the water + sample = $43.0\text{mL}$

Then,

Volume of the sample = 43.0 − 40.0

= 3.0 mL

$\begin{array}{c}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ =\frac{20.46\text{ g}}{3.0\text{ mL}}\\ =6.82\text{g/mL}\end{array}$

The error associated with the experimental value is the difference of the experimental and accepted value.

$\begin{array}{c}\text{Error}=\text{experimental}\text{value}-\text{accepted}\text{value}\\ =6.82-6.95\\ =-0.13\end{array}$

Percent error represents the error as the percentage of the accepted value.

$\begin{array}{c}\text{Percent}\text{error}=\frac{\left|\text{error}\right|}{\text{accepted}\text{value}}\times 100\\ =\frac{\left|-0.13\right|}{\text{6}\text{.95}}\times 100\\ =\frac{0.13}{\text{6}\text{.95}}\times 100\\ =1.87\%\end{array}$

Conclusion

Thus, the density and the percent error of the sample is 6.82 g/mL and 1.87% respectively.