To find the mass of the spherical ball of lead, we can use the formula:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
First, we need to calculate the volume of the sphere. The formula for the volume \( V \) of a sphere is:
\[ V = \frac{4}{3} \pi r^3 \]
Given that the diameter of the sphere is 5.0 cm, we can find the radius \( r \):
\[ r = \frac{\text{diameter}}{2} = \frac{5.0 , \text{cm}}{2} = 2.5 , \text{cm} \]
Now, we can calculate the volume:
\[ V = \frac{4}{3} \pi (2.5 , \text{cm})^3 \]
Calculating \( (2.5 , \text{cm})^3 \):
\[ (2.5)^3 = 15.625 , \text{cm}^3 \]
Now substituting back into the volume formula:
\[ V = \frac{4}{3} \pi (15.625 , \text{cm}^3) \]
Using \( \pi \approx 3.14159 \):
\[ V \approx \frac{4}{3} \times 3.14159 \times 15.625 \approx 65.45 , \text{cm}^3 \]
Next, we will calculate the mass using the density of lead, which is given as 11.34 g/cm³:
\[ \text{Mass} = \text{Density} \times \text{Volume} = 11.34 , \text{g/cm}^3 \times 65.45 , \text{cm}^3 \]
Calculating the mass:
\[ \text{Mass} \approx 11.34 \times 65.45 \approx 742.78 , \text{g} \]
Finally, rounding this to two significant figures, we get:
\[ \text{Mass} \approx 740 , \text{g} \]
Therefore, the mass of the sphere is approximately 740 g.