# Question 6f827

Sep 26, 2015

$\text{1.57 g/mL}$

#### Explanation:

The idea here is that the density of the mixture will reflect the contribution the volume of each substance has in the total volume of the mixture.

In your case, you have equal volumes of butter and of sand, which means that the density of the mixture will be split half way between the two densities of the components of the mixture.

Mathemcatically, you can prove this using

${\rho}_{1} = {m}_{1} / V \to$ the density of butter

and

${\rho}_{2} = {m}_{2} / V \to$ the density of sand

The volume $V$ is the same for both substances, since you're mixing $\text{1.00 mL}$ of each.

Now, the density of the mixture will be

${\rho}_{\text{mixture" = m_"total"/V_"total" " }}$, where

${m}_{\text{total}}$ - the total mass of the mixture, equal to ${m}_{1} + {m}_{2}$;
${V}_{\text{total}}$ - the total volume of the mixture, equal to $V + V$

This means that you have

${\rho}_{\text{mixture}} = \frac{{m}_{1} + {m}_{2}}{V + V} = \frac{{m}_{1} + {m}_{2}}{2 V}$

${\rho}_{\text{mixture}} = \frac{1}{2} \cdot \frac{{m}_{1} + {m}_{2}}{V} = \frac{1}{2} \cdot \left({m}_{1} / V + {m}_{2} / V\right)$

But ${m}_{1} / V = {\rho}_{1}$ and ${m}_{2} / V = {\rho}_{2}$, so you have

${\rho}_{\text{mixture}} = \frac{1}{2} \cdot \left({\rho}_{1} + {\rho}_{2}\right)$

Therefore,

rho_"mixture" = 1/2 * (0.860 + 2.28)"g/mL" = color(green)("1.57 g/mL")#