Density

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MCAT Physical › Density

Questions 1 - 10

An object is at equilibrium when of the total volume is submerged in gasoline. Find the density of the object.

$\small \rho_{gasoline} = 0.8\frac{g}{mL}$

$0.264\frac{g}{mL}$

$0.8\frac{g}{mL}$

$0.736\frac{g}{mL}$

$0.264\frac{g}{L}$

$0.736\frac{g}{L}$

Explanation

To solve, we can equate the volume of the gasoline displaced to the volume of the portion of the object that is submerged.

$V_{gas}=V_{obj}$

$V=\frac{m}{\rho}$

We know that of the object is submerged, thus of the object's total volume will equal the volume of water displaced.

$\frac{m_{gas}}{\rho_{gas}}=0.33*\frac{m_{obj}}{\rho_{obj}}$

$\frac{\rho_{obj}}{\rho_{gas}}=0.33\frac{m_{obj}}{m_{gas}}$

The displaced masses are equal.

$\frac{\rho_{obj}}{\rho_{gas}}=0.33$

$\rho_{obj}=(0.33)(0.8\frac{g}{mL})=0.264\frac{g}{mL}$

A researcher performs an elemental analysis on a compound. He finds that the compound is made up of only carbon, hydrogen, and oxygen atoms. He isolates a pure sample of the compound and finds that this sample contains of carbon, of hydrogen, and of oxygen. The researcher wants to perform further analysis on this compound the next day. Before leaving the lab the researcher creates three stock solutions of varying concentrations of this compound: (solution A), (solution B), and (solution C). He stores these solutions overnight at a temperature of .

Molecular weight of this compound = $300.486\frac{g}{mol}$

Which of the following is true regarding the densities of the three stock solutions?

$\rho_A<\rho_B<\rho_C$

$\rho_A=\rho_B=\rho_C$

$\rho_A>\rho_B>\rho_C$

$\rho_A=\rho_B<\rho_C$

Explanation

Density is a measure of concentration, defined as mass per unit volume.

$\rho=\frac{\text{mass}}{\text{volume}}$

Since the stock solutions contain different concentrations, the density of each solution must be different. The solution with the highest concentration will contain the highest density, and the solution with lowest concentration will contain the lowest density. The order of increasing density is solution A, solution B, and solution C. Solution C is the most concentrated and will contain the greatest amount of mass per unit volume, followed by solution B and solution A.

$\rho_A<\rho_B<\rho_C$

Remember that concentration can be defined in various ways: molarity, molality, mass percent, density, etc. Increasing concentration means you are increasing all of these quantities. Also remember that the density of the compound remains the same in each solution; only the solutions, as a whole, have different densities.

Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.

Phase_diagram_ps

The density of the solution in section 1 is __________ the density of the solution in section 2.

less than

greater than

equal to

Explanation

The slope of the solid/liquid phase transition line can predict the comparative density of the solution in each section. A negative slope indicates that the solid phase is less dense than the liquid phase. A positive slope indicates that the solid is more dense. In the above phase diagram, the slope is negative, indicating that the solid is less dense than the liquid.

$\rho_{H_2O}=1.0\frac{g}{mL}$

$\rho_{Al}=2.7\frac{g}{mL}$

$\rho_{Pb}=11.4\frac{g}{mL}$

Which of these samples, if any, has a volume greater than three liters?

$19lbs\ Al$

$6.0lbs H_{2}O$

$50lbs\ Pb$

$19lbs\ Al$

Both the aluminum and water samples

$50lbs\ Pb$

None of these has a volume greater than three liters

Explanation

Solving this question requires that we use the given densities to convert mass to volume.

$6.0 lb H_{2}O * \frac{454 g}{lb} * \frac{1 mL}{1.0 g}* \frac{1 L}{1000 mL} = 2.7 L$

$50 lb Pb * \frac{454 g}{lb} * \frac{1 mL}{11.4 g}* \frac{1 L}{1000 mL} = 2.0 L$

$19 lb Al * \frac{454 g}{lb} * \frac{1 mL}{2.7 g}* \frac{1 L}{1000 mL} = 3.2 L$

We see that only the aluminum sample has a volume over three liters.

Solid A has a volume of $36:cm^{3}$ and a density of $0.25:\frac{g}{cm^{3}}$ . Solid B is cube with sides of $\small 5cm$ and has a density of $0.10:\frac{g}{cm^{3}}$ .

What is the difference in mass between the two solids?

Explanation

The formula for density is:

$\rho=\frac{mass}{volume}$

In the question, we are given the densities of both solids and a means to find their volumes. Using these values, we will be able to determine the mass of each solid.

$m=\rho*V$

$m_A=(0.25\frac{g}{cm^3})(36cm^3)=9g$

$m_B=(0.10\frac{g}{cm^3})(5cm)^3=12.5g$

Now that we know both masses, we can find the difference.

What is the density of a log that is 25cm long, has a cross sectional area of 5cm2, and weighs 100g?

$0.80\frac{g}{cm^3}$

$3.0\frac{g}{cm^3}$

$15.0\frac{g}{cm^3}$

$20.0\frac{g}{cm^3}$

$0.25\frac{g}{cm^3}$

Explanation

The density of an object is equal to mass over volume.

$\rho=\frac{m}{V}$

Knowing the length and the cross sectional area of the log, we can find its volume.

Plugging in volume and mass into the equation will enable us to find density.

$\rho=\frac{100g}{125cm^3}=0.80\frac{g}{cm^3}$

Water has a density of $\small 1.00\frac{g}{cm^3}$ at a temperature of $\small 4^oC$ . An certain oil has a specific gravity of $\small 0.882$ at this temperature. What is the mass of $\small 1L$ of the oil?

Explanation

The formula for density is:

$\rho=\frac{\text{mass}}{\text{volume}}=\frac{m}{V}$

Specific gravity is the density of a material relative to the density of water.

$SG_{oil}=\frac{\rho_{oil}}{\rho_{water}}$

We are given the specific gravity of the oil and the density of water, allowing us to calculate the density of the oil.

$0.882=\frac{\rho_{oil}}{1.00\frac{g}{cm^3}}$

$\rho_{oil}=(0.882)(1.00\frac{g}{cm^3})=0.882\frac{g}{cm^3}$

Returning to the equation for density, we can use the density of the oil to find the mass on one liter.

$\rho_{oil}=\frac{m}{V}$

$0.882\frac{g}{cm^3}=\frac{m}{1000cm^3}$

$m=(0.882\frac{g}{cm^3})(1000cm^3)=882g$

A given iceberg floats such that 90% of its volume is below water. Supposing that the iceberg is composed of pure water, what is the density of the saltwater in which the iceberg floats?

$\rho_{ice}=0.917\frac{g}{mL}$

$1.019\frac{g}{mL}$

$1.000\frac{g}{mL}$

$0.9017\frac{g}{mL}$

$0.7063\frac{g}{mL}$

$1.112\frac{g}{mL}$

Explanation

We can solve this question by using a density ratio, given by the equation:

$X=\frac{\rho_{iceberg}}{\rho_{saltwater}}$

We know the density of ice, and we can find the ratio of the densities by using buoyancy. If the iceberg were completely submerged, then we could conclude that the density of the ice was equal to the density of the salt water. Since the iceberg is 90% submerged, we can conclude that the density of the ice is 90% of the density of the salt water.

$\frac{\rho_{iceberg}}{\rho_{saltwater}}=0.9$

Use this ratio and the density of ice to solve for the density of the salt water.

$0.9=\frac{0.917\frac{g}{mL}}{\rho_{saltwater}}$

$\rho_{saltwater}=1.019\frac{g}{mL}$

Molecular weight of this compound = $300.486\frac{g}{mol}$

A stock solution, solution X, kept at room temperature () will have __________ compared to solution A.

a lower density

a greater density

the same density

Relative densities can’t be determined without more information

Explanation

Solution A and solution X have the same concentration, therefore, we are only concerned with temperature differences between the two solutions. Density is dependent on temperature: as temperature increases density decreases. Recall the definition of density:

$\rho=\frac{\text{mass}}{\text{volume}}$

Increasing the temperature will slightly increase the volume of the solution and, subsequently, decrease density. The temperature has no effect on mass. The solution at the higher temperature (solution X) will have a lower density.

What is the specific gravity of a boat that has a mass of 6000kg and a volume of 10m3?

0.6