**Mass **(m) is equal to the product of **density **(ρ) and the **volume **(v). Using the formula of mass: **m = ρ × v**, the value of mass of an object can be calculated.

Let’s solve some problems based on this formula, so you’ll get a clear idea.

## Mass Practice Problems

**Problem 1:** Calculate the mass of a steel, if the density of a steel is 8050 kg/m^{3} and the volume of a steel is 0.012 m^{3}.

Solution:

Given data:

Mass of a steel, m = ?

Density of a steel, ρ = 8050 kg/m^{3}

Volume of a steel, v = 0.012 m^{3}

Using the formula of mass,

m = ρ × v

m = 8050 × 0.012

m = 96.6 kg

Therefore, the mass of a steel is **96.6 kg**.

**Problem 2:** If the density of a diamond is 3510 kg/m^{3} and the volume of a diamond is 0.005 m^{3}, then calculate the mass of a diamond.

Solution:

Given data:

Density of a diamond, ρ = 3510 kg/m^{3}

Volume of a diamond, v = 0.005 m^{3}

Mass of a diamond, m = ?

Using the formula of mass,

m = ρ × v

m = 3510 × 0.005

m = 17.55 kg

Therefore, the mass of a diamond is **17.55 kg**.

**Problem 3:** The density of a silver is 10497 kg/m^{3} and the volume of a silver is 0.008 m^{3}. Calculate the mass of a silver.

Solution:

Given data:

Density of a silver, ρ = 10497 kg/m^{3}

Volume of a silver, v = 0.008 m^{3}

Mass of a silver, m = ?

Using the formula of mass,

m = ρ × v

m = 10497 × 0.008

m = 83.97 kg

Therefore, the mass of a silver is **83.97 kg**.

**Problem 4:** Find the mass of an ice having a density 917 kg/m^{3} and the volume 0.007 m^{3}.

Solution:

Given data:

Mass of ice, m = ?

Density of ice, ρ = 917 kg/m^{3}

Volume of ice, v = 0.007 m^{3}

Using the formula of mass,

m = ρ × v

m = 917 × 0.007

m = 6.41 kg

Therefore, the mass of an ice is **6.41 kg**.

.

.

.

**Related:**

- Free Fall Equation
- Air Resistance Formula
- Terminal Velocity Equation
- Gravity Equation
- Inertia Formula
- Acceleration Formula
- Momentum Equation
- Velocity Formula
- Pressure Equation
- Kinematic Equations
- Friction Equation

**Also Read:**