In a particuar amount of steel, there `80%` bronze and `20%` zinc. Density of bronze is `8.9g//cm^(3)` and that of zinc is `7.1cm//cm^(3)`. Find the density of steel.

Options

A. `8.5g//cm^(3)` (approx.)
B. `8.0g//cm^(3)` (approx.)
C. `7.9g//cm^(3)` (approx.)
D. `8.9g//cm^(3)` (approx.)

Answer

To find the density of steel composed of 80% bronze and 20% zinc, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( V \) = total volume of the steel
- Density of bronze, \( D_B = 8.9 \, \text{g/cm}^3 \)
- Density of zinc, \( D_Z = 7.1 \, \text{g/cm}^3 \)

### Step 2: Calculate the Volume of Each Component
Since 80% of the volume is bronze and 20% is zinc:
- Volume of bronze, \( V_B = 0.8V \)
- Volume of zinc, \( V_Z = 0.2V \)

### Step 3: Calculate the Mass of Each Component
Using the formula for mass (\( \text{mass} = \text{density} \times \text{volume} \)):
- Mass of bronze, \( M_B = D_B \times V_B = 8.9 \, \text{g/cm}^3 \times (0.8V) = 7.12V \, \text{g} \)
- Mass of zinc, \( M_Z = D_Z \times V_Z = 7.1 \, \text{g/cm}^3 \times (0.2V) = 1.42V \, \text{g} \)

### Step 4: Calculate the Total Mass
Total mass of the steel, \( M \):
\[ M = M_B + M_Z = 7.12V + 1.42V = 8.54V \, \text{g} \]

### Step 5: Calculate the Density of Steel
Using the formula for density (\( \text{density} = \frac{\text{mass}}{\text{volume}} \)):
\[ D_{steel} = \frac{M}{V} = \frac{8.54V}{V} = 8.54 \, \text{g/cm}^3 \]

### Conclusion
The density of the steel is approximately \( 8.54 \, \text{g/cm}^3 \).

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