Question

A metal slug weighing \(25.17 \mathrm{~g}\) is added to a flask with a volume of \(59.7 \mathrm{~mL}\). It is found that \(43.7 \mathrm{~g}\) of methanol \((d=0.791 \mathrm{~g} / \mathrm{mL})\) must be added to the metal to fill the flask. What is the density of the metal?

Short answer

Answer: The density of the metal slug is 5.65 g/mL.

Step-by-step solution

Calculate the volume of methanol

To calculate the volume of methanol needed to fill the flask, we will use the given mass of methanol and its density. We can use the formula: Volume = (Mass of methanol) / (Density of methanol) \(V_\mathrm{methanol} = \frac{43.7 \mathrm{~g}}{0.791 \mathrm{~g} / \mathrm{mL}} = 55.25 \mathrm{~mL}\)

Calculate the volume of the metal slug

Now, we can find the volume of the metal slug by subtracting the volume of methanol from the volume of the flask. Volume of metal slug = Volume of flask - Volume of methanol \(V_\mathrm{metal} = 59.7 \mathrm{~mL} - 55.25 \mathrm{~mL} = 4.45 \mathrm{~mL}\)

Calculate the density of the metal slug

Having found the volume of the metal slug, we can now find its density using the given mass and the calculated volume. We can use the formula: Density = (Mass of the metal slug) / (Volume of the metal slug) \(\rho_\mathrm{metal} = \frac{25.17 \mathrm{~g}}{4.45 \mathrm{~mL}} = 5.65 \mathrm{~g} / \mathrm{mL}\) So, the density of the metal slug is \(5.65 \mathrm{~g} / \mathrm{mL}\).