If the specific gravity of a liquid is 0.76, calculate its density(($\rho_w$ = 1000Kgm$^{-3}$)

1. Analyze the question

You are asked to find the density of a liquid given its specific gravity (also known as relative density) and the density of water.

2. Explain the concept

Specific Gravity (Relative Density) is a dimensionless quantity defined as the ratio of the density of a substance to the density of a reference substance (usually water).

The formula is:

To find the density of the substance, we rearrange the formula:

3. Show the solution

• Given: • Specific Gravity = 0.76 • Density of water ($\rho_w$) = 1000 kgm⁻³

• Calculation: $$\text{Density of liquid} = 0.76 \times 1000 \text{ kgm}^{-3}$$ $$\text{Density of liquid} = 760 \text{ kgm}^{-3}$$

4. Identify the answer

Comparing our result to the given options: A. 760 kgm⁻³ B. 76 kgm⁻³ C. 7.6 kgm⁻³ D. 7600 kgm⁻³

The correct option is A.

5. Tip

When dealing with specific gravity, remember that it is just a multiplier. If you are converting from specific gravity to density in SI units (kgm⁻³), simply multiply by 1000. If you are converting to CGS units (gcm⁻³), the specific gravity value is numerically equal to the density (e.g., 0.76 gcm⁻³). This is a quick way to verify your work during the exam!