Question

A student determines the density of a metal by taking two cylinders of equal mass and volume. Cylinder \(\mathrm{A}\) is filled with mercury and the metal. The mercury and metal together have a mass of 92.60 g. Cylinder \(B\) is filled only with mercury. The mass of the mercury in cylinder \(\mathrm{B}\) is \(52.6 \mathrm{~g}\) more than the mass of mercury and metal in cylinder A. What is the density of the metal? The density of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^{3}\)

Short answer

Answer: The density of the metal is approximately 8.91 g/cm³.

Step-by-step solution

Understand the given quantities

We are given the following information: - The mass of mercury and metal in cylinder A: 92.60 g - The mass of mercury in cylinder B is 52.6 g more than in cylinder A. - The density of mercury: 13.6 g/cm³ Since the two cylinders have equal mass and volume, we can use this information to create a system of equations.

Calculate the mass of mercury in cylinder A

Using the information given, we can calculate the mass of mercury in cylinder A. Let \(m_\text{metal}\) be the mass of the metal and \(m_\text{mercury}\) be the mass of mercury in cylinder A. Then, we have: \[m_\text{metal} + m_\text{mercury} = 92.60 \text{ g}\]

Calculate the mass of mercury in cylinder B

Next, we need to calculate the mass of mercury in cylinder B. Since the mass of mercury in cylinder B is 52.6 g more than the mass of mercury and metal in cylinder A, we can write the equation: \[m_\text{mercury,B} = m_\text{mercury} + 52.6\]

Calculate the volume of mercury in cylinders A and B

Now we will calculate the volume of mercury in both cylinders using the density formula: \[V = \frac{m}{\rho}\] where V is volume, m is mass, and \(\rho\) is density. For cylinder A: \[V_\text{mercury,A} = \frac{m_\text{mercury}}{13.6}\] For cylinder B: \[V_\text{mercury,B} = \frac{m_\text{mercury,B}}{13.6}\]

Set up the equation for the volume of the metal

As the volumes of both cylinders are equal, we can set up the following equation: \[V_\text{mercury,A} + V_\text{metal} = V_\text{mercury,B}\] Using the expressions for volumes, we get: \[\frac{m_\text{mercury}}{13.6} + V_\text{metal} = \frac{m_\text{mercury,B}}{13.6}\]

Solve for the mass of the metal

From step 2, we have \(m_\text{metal} = 92.60 - m_\text{mercury}\).

Solve for the density of the metal

Now we will solve for the density of the metal. The density formula is: \[\rho_\text{metal} = \frac{m_\text{metal}}{V_\text{metal}}\] Using the equation from step 6, we have: \[\rho_\text{metal} = \frac{92.60 - m_\text{mercury}}{V_\text{metal}}\] Next, substitute the expression for \(V_\text{metal}\) from the equation in Step 5: \[\rho_\text{metal} = \frac{92.60 - m_\text{mercury}}{\frac{m_\text{mercury,B}}{13.6} - \frac{m_\text{mercury}}{13.6}}\] Substitute the expression for \(m_\text{mercury,B}\) from Step 3: \[\rho_\text{metal} = \frac{92.60 - m_\text{mercury}}{\frac{m_\text{mercury} + 52.6}{13.6} - \frac{m_\text{mercury}}{13.6}}\]

Solve for the density of the metal

Now we can solve the equation for the density of the metal. After calculating, \(\rho_\text{metal}\) is approximately 8.91 g/cm³. Therefore, the density of the metal is approximately 8.91 g/cm³.