Q3) A. If a 70% (by weight) solution of glycerol has a specific gravity of 1.184 at 15°C, what is the density of the solution in (a) g/cm³? (b) Ibm/ft³? and (c) kg/m³?
The Correct Answer and Explanation is :
To determine the density of a 70% (by weight) glycerol solution with a specific gravity of 1.184 at 15°C in various units, we can use the relationship between specific gravity and density. Specific gravity is defined as the ratio of the density of a substance to the density of a reference substance, typically water at 4°C, which has a density of 1 g/cm³ (1000 kg/m³ or 62.43 lb/ft³).
(a) Density in g/cm³:
Given that the specific gravity (SG) is 1.184, the density (ρ) in g/cm³ can be calculated as:
[ \rho = \text{SG} \times \text{Density of water} ]
Since the density of water is 1 g/cm³:
[ \rho = 1.184 \times 1\, \text{g/cm}^3 = 1.184\, \text{g/cm}^3 ]
(b) Density in lbm/ft³:
To convert the density from g/cm³ to lbm/ft³, we use the conversion factor:
[ 1\, \text{g/cm}^3 = 62.428\, \text{lbm/ft}^3 ]
Therefore:
[ \rho = 1.184\, \text{g/cm}^3 \times 62.428\, \text{lbm/ft}^3 = 73.914752\, \text{lbm/ft}^3 ]
(c) Density in kg/m³:
Since 1 g/cm³ is equivalent to 1000 kg/m³:
[ \rho = 1.184\, \text{g/cm}^3 \times 1000 = 1184\, \text{kg/m}^3 ]
Explanation:
Specific gravity is a dimensionless quantity that expresses the ratio of the density of a substance to the density of a reference substance, typically water at its maximum density (4°C). This property allows for the easy comparison of densities without the need for units. In this case, a specific gravity of 1.184 indicates that the glycerol solution is 1.184 times denser than water.
To find the density in g/cm³, we multiply the specific gravity by the density of water in the same units. Since the density of water is 1 g/cm³, the calculation is straightforward, yielding 1.184 g/cm³.
For the density in lbm/ft³, we use the fact that 1 g/cm³ equals 62.428 lbm/ft³. Multiplying the density in g/cm³ by this conversion factor provides the density in lbm/ft³.
Similarly, to express the density in kg/m³, we recognize that 1 g/cm³ is equivalent to 1000 kg/m³. Thus, multiplying the density in g/cm³ by 1000 gives the density in kg/m³.
Understanding these conversions is crucial in fields like chemical engineering and material science, where precise measurements of substance properties are essential for process design, safety assessments, and quality control. Accurate density values influence calculations involving mass and volume, affecting the design and operation of equipment, mixing processes, and storage requirements.