Chapter 7, Problem 124A

Interpretation Introduction

Interpretation:

The density of the sample and the percent error if the accepted value of this sample is 7.01 g/mL needs to be determined.

Concept introduction:

Density can be defined as the mass per unit volume. The symbol $\rho$ is used for density. The unit of density can be expressed in g/mL

$\rho =\frac{m}{v}$

Answer to Problem 124A

The density of the sample is $4.685\text{g/mL}$ and the percentage error is 49.78 %.

Explanation of Solution

Given,

Mass of the sample is = 19.21 g

Volume of water + sample is = 43.1 mL

Volume of the water alone is = 39.0 mL

So, the volume of water displaced = 43.1 mL - 39.0 mL = 4.1 mL

Therefore, the density of the sample,

$\begin{array}{l}\rho =\frac{m}{v}\\ =\frac{massofthesample}{\begin{array}{l}volumeofthewaterdisplaced\\ =\frac{19.21\text{g}}{4.1\text{mL}}\\ =4.685\text{g/mL}\end{array}}\end{array}$

The percent error if the accepted value of this sample is 7.01 g/mL can be calculated using the value, $\text{(7}\text{.01-4}\text{.68)}\text{g/mL=}\text{2}\text{.33}\text{g/mL}$

$\begin{array}{c}\text{Percentage error}\text{=}\frac{\text{2}\text{.33}\text{g/mL}}{\text{4}\text{.68}\text{g/mL}}\times 100\\ =49.78\%\end{array}$

Conclusion

From the given data, we have calculated thedensity of the sample to be 4.685 g/mL and the percentage error is 49.78 %.