Chapter 5, Problem 70A

Interpretation Introduction

Interpretation : The total mixture of the solution of benzene and ethanol is to be calculated.

Concept Introduction :

The density of the substance is the mass per unit volume. It is given as,

  $\rho =\frac{m}{V}$

Where, $\rho$ is the density of the object

  $m$ is the mass of the object

  $V$ is the volume of the object

Answer to Problem 70A

The total mass of the mixture is $\text{958 g}$ .

Explanation of Solution

Given information :

Volume of the ethanol, $V_{\text{ethanol}}\text{=}\text{100}\text{.0 mL}$

Volume of the benzene, $V_{\text{benzene}}\text{=}\text{1L=1000mL}$

Density of the ethanol, ${\rho }_{\text{ethanol}}=0.78\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{\text{3}}$}\right.$

Density of the benzene, ${\rho }_{\text{benzene}}\text{=}\text{0}\text{.880}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{\text{3}}$}\right.$

And,

  $\text{1mL=}{\text{1cm}}^{\text{3}}$

Calculation: The density formula is represented as follows:

  $\rho =\frac{m}{V}$

Firstly, calculate the mass of the ethanol by substituting the given values, such that,

  $\begin{array}{c}\rho =\frac{m}{V}\\ m=V_{\text{ethanol}}\times {\rho }_{\text{ethanol}}\\ m_{\text{ethanol}}=100.\text{0mL×}\text{0}\text{.78}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{\text{3}}$}\right.\\ \text{=78g}\end{array}$

Similarly, calculate the mass of the benzene,

  $\begin{array}{c}\rho =\frac{m}{V}\\ m=V_{\text{benzene}}\times {\rho }_{\text{benzene}}\\ m_{benzene}=\text{1000 mL×}\text{0}\text{.880}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{\text{3}}$}\right.\\ \text{=880 g}\end{array}$

ow, the total mass of the mixture is,

  $m_{\text{mixture}}=m_{\text{ethanol}}+m_{\text{benzene}}$

Now, substitute the values in this above equation, then

  $\begin{array}{l}{m}_{\text{mixture}}=m_{\text{ethanol}}+m_{\text{benzene}}\\ \text{=}\text{78g+880g}\\ \text{ =}\text{958g}\end{array}$

Conclusion

The mass of the solution after mixing the ethanol into benzene is $\text{958 g}$ .