Question

1. use the data table below and the attached graph paper to plot the mass and volume of the 5 samples of the mineral galena below. note: the resulting line you plot is the mineral’s density!
2. calculate the density of samples 1 - 5 and place the value in the “density” column of the data table below.
3. answer the questions below.

sample	size	mass	volume	density (d = m/v)
2		60 g	8 cm³
3		120 g	16 cm³
4		480 g	64 cm³
5	large	750 g	100 cm³

questions

1. use the completed graph to determine how much mass a sample of galena would have if its volume was 75 cm³.
2. use the completed graph to determine how much volume a sample of galena would have if its mass was 300 g.
3. describe the relationship between mass and volume shown by the graph.
4. how does the density of sample 2 compare to that of sample 4?
5. how did the density of the largest sample (5) compare to the smallest sample (1)?
6. what is the effect of sample size on the density of a material?
7. describe the trick you were taught on how to change around the density formula to solve for different parts of it.

Explanation:

Step1: Calculate density for Sample 1

Density formula: $d = \frac{m}{v}$. For Sample 1, $m = 15\ g$, $v = 2\ cm^3$. So $d_1 = \frac{15}{2} = 7.5\ g/cm^3$.

Step2: Calculate density for Sample 2

For Sample 2, $m = 60\ g$, $v = 8\ cm^3$. $d_2 = \frac{60}{8} = 7.5\ g/cm^3$.

Step3: Calculate density for Sample 3

For Sample 3, $m = 120\ g$, $v = 16\ cm^3$. $d_3 = \frac{120}{16} = 7.5\ g/cm^3$.

Step4: Calculate density for Sample 4

For Sample 4, $m = 480\ g$, $v = 64\ cm^3$. $d_4 = \frac{480}{64} = 7.5\ g/cm^3$.

Step5: Calculate density for Sample 5

For Sample 5, $m = 750\ g$, $v = 100\ cm^3$. $d_5 = \frac{750}{100} = 7.5\ g/cm^3$.

Question 1:

From the graph (assuming linear relationship as density is constant), since $d = 7.5\ g/cm^3$, for $v = 75\ cm^3$, $m = d\times v = 7.5\times75 = 562.5\ g$.

Question 2:

For $m = 300\ g$, $v = \frac{m}{d} = \frac{300}{7.5} = 40\ cm^3$.

Question 3:

The graph shows a direct proportional relationship (linear, passing through origin) between mass and volume, as $m = d\times v$ (density constant), so mass increases linearly with volume.

Question 4:

Density of Sample 2 ($d_2 = 7.5\ g/cm^3$) is equal to density of Sample 4 ($d_4 = 7.5\ g/cm^3$) because density is an intensive property (depends on substance, not sample size).

Question 5:

Density of largest sample (5, $d_5 = 7.5\ g/cm^3$) is equal to density of smallest sample (1, $d_1 = 7.5\ g/cm^3$) as density is substance - specific, not dependent on sample size.

Question 6:

Sample size (amount of substance) does not affect the density of a material. Density is an intensive property, so it remains constant for a pure substance regardless of how much of the substance is present.

Question 7:

The density formula is $d=\frac{m}{v}$. To solve for mass, rearrange to $m = d\times v$ (multiply both sides by $v$). To solve for volume, rearrange to $v=\frac{m}{d}$ (divide both sides by $d$). A common trick is to use the "triangle" method: draw a triangle with $m$ at the top, and $d$ and $v$ at the bottom corners. To find $m$, cover $m$ and multiply $d$ and $v$; to find $d$, cover $d$ and divide $m$ by $v$; to find $v$, cover $v$ and divide $m$ by $d$.

Answer:

s:

1. The mass would be $\boldsymbol{562.5\ g}$.
2. The volume would be $\boldsymbol{40\ cm^3}$.
3. Mass and volume have a direct proportional (linear) relationship (mass increases linearly with volume, $m = k\times v$ where $k$ is density).
4. The density of Sample 2 is equal to the density of Sample 4.
5. The density of the largest sample (5) is equal to the density of the smallest sample (1).
6. Sample size has no effect on the density of a material (density is an intensive property, constant for a pure substance).
7. Use the density formula $d=\frac{m}{v}$ and rearrange: $m = d\times v$ (for mass), $v=\frac{m}{d}$ (for volume). A "triangle" trick (cover the desired quantity: multiply for $m$, divide for $d$ or $v$) helps.