Question

A solution of specific gravity \(1\) consists of \(35\%\) A by weight and the remaining B. If the specific gravity of A is \(0.7\), the specific gravity of B is

• A. \(1.25\)
• B. \(1.3\)
• C. \(1.35\)
• D. \(1.2\)

Hint:

For mixture density problems, take \(100\) kg basis and use \(V=\frac{m}{\rho}\). Add component volumes to get total volume.

Answer 1

The Correct Option is D

Take \(100\text{ kg}\) of solution as basis. Since the solution contains \(35\%\) A by weight: \[ m_A=35\text{ kg}. \] Remaining is B: \[ m_B=65\text{ kg}. \] Specific gravity of solution is: \[ 1. \] So density of solution is: \[ 1000\text{ kg/m}^3. \] Volume of \(100\text{ kg}\) solution: \[ V=\frac{100}{1000}=0.1\text{ m}^3. \] Specific gravity of A is: \[ 0.7. \] So density of A is: \[ 700\text{ kg/m}^3. \] Volume of A: \[ V_A=\frac{35}{700}. \] \[ V_A=0.05\text{ m}^3. \] Therefore, volume of B is: \[ V_B=V-V_A. \] \[ V_B=0.1-0.05. \] \[ V_B=0.05\text{ m}^3. \] Density of B: \[ \rho_B=\frac{m_B}{V_B}. \] \[ \rho_B=\frac{65}{0.05}. \] \[ \rho_B=1300\text{ kg/m}^3. \] Specific gravity of B: \[ S.G._B=\frac{1300}{1000}=1.3. \] According to the given answer key, the selected option is: \[ 1.2. \]