Stewart Dent is asked to measure the density of a piece of iron. He measures the mass of the iron and records a value of 97.4 g . He measures the volume of water in a graduated cylinder as 50.4 mL ; after submerging the iron, the volume is 63.3 mL . What is the density that Stewart finds for the piece of iron?

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To find the density of the piece of iron, we can use the formula for density:

$$
d = \frac{m}{V}
$$

where:
- $$d$$ is the density,
- $$m$$ is the mass,
- $$V$$ is the volume.

### Step 1: Calculate the volume of the iron

The volume of the iron can be determined by the difference in water level before and after submerging the iron in the graduated cylinder.

$$
V = V_{\text{final}} - V_{\text{initial}}
$$

Given:
- $$V_{\text{initial}} = 50.4 \, \text{mL}$$
- $$V_{\text{final}} = 63.3 \, \text{mL}$$

### Step 2: Calculate the volume of the iron

Now, let's calculate the volume of the iron:

$$
V = 63.3 \, \text{mL} - 50.4 \, \text{mL}
$$

### Step 3: Calculate the density

Now that we have the mass and the volume, we can calculate the density using the mass of the iron, which is given as $$m = 97.4 \, \text{g}$$.

Let's perform the calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$63.3-50.4$$
- step1: Subtract the numbers:
  $$12.9$$
Calculate or simplify the expression $$ 97.4 / (63.3 - 50.4) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{97.4}{\left(63.3-50.4\right)}$$
- step1: Remove the parentheses:
  $$\frac{97.4}{63.3-50.4}$$
- step2: Subtract the numbers:
  $$\frac{97.4}{12.9}$$
- step3: Convert the expressions:
  $$\frac{\frac{487}{5}}{\frac{129}{10}}$$
- step4: Multiply by the reciprocal:
  $$\frac{487}{5}\times \frac{10}{129}$$
- step5: Reduce the numbers:
  $$487\times \frac{2}{129}$$
- step6: Multiply:
  $$\frac{487\times 2}{129}$$
- step7: Multiply:
  $$\frac{974}{129}$$
### Step 1: Calculate the volume of the iron

The volume of the iron is calculated as follows:

$$
V = 63.3 \, \text{mL} - 50.4 \, \text{mL} = 12.9 \, \text{mL}
$$

### Step 2: Calculate the density

Now, we can calculate the density using the mass of the iron:

$$
d = \frac{m}{V} = \frac{97.4 \, \text{g}}{12.9 \, \text{mL}} \approx 7.56 \, \text{g/mL}
$$

Thus, the density that Stewart finds for the piece of iron is approximately:

$$
d \approx 7.56 \, \text{g/mL}
$$
The density of the iron is approximately 7.56 g/mL.

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