Question

Part BIn liposuction, a doctor removes fat deposits from a person's body. If body fat has a density of 0.909 g/mL and 3.8 L of fat is removed, how many pounds of fat are removed from the patient? Express your answer using two significant figures.

Answer 1

The calculation for the weight of fat removed in liposuction from 3.8 liters of body fat (density 0.909 g/mL) results in approximately 7.6 pounds of fat.

Step-by-step explanation:

To calculate how many pounds of fat are removed in a liposuction procedure, where 3.8 liters of fat with a density of 0.909 g/mL is extracted, we can follow these steps:

Convert liters to milliliters: 3.8 L × 1000 mL/L = 3800 mL.

Multiply the volume by the density to get the mass in grams: 3800 mL × 0.909 g/mL = 3454.2 g.

Convert grams to pounds (1 pound = 453.59237 grams): 3454.2 g × (1 pound / 453.59237 g) ≈ 7.62 pounds.

So, approximately 7.6 pounds of fat are removed during the procedure.

Answer 2

First, we have to remember that density is a property that matches the mass and the volume of a determined substance.

`\rho=\text{ }(mass)/(volume)`

If we have the volume and the density of the fat deposits, we can calculate the mass by clearing the equation as follows:

`\begin{gathered} volume\text{ * }\rho\text{ = mass} \\ \\ 3.8\text{ L *}\frac{1000\text{ mL}}{1\text{ L}}*\text{ 0.909 }(g)/(mL)=\text{ 3454.2 g } \\ \end{gathered}`

And now, we can calculate the pounds by using a conversion factor:

`3454.2\text{ g *}\frac{1\text{ lb}}{453.592\text{ g}}=\text{ 7.6152 lb }\approx\text{ 7.6 lb}`

Then, the removed mass of fat is 7.6 lb approx.