Mixture specific gravity calculation:\nA solution has overall specific gravity 1.000 and contains 35% component A by weight (the balance is component B). If the specific gravity of pure A is 0.70, determine the specific gravity of pure B.

Introduction / Context:
Process engineers often estimate unknown component densities in liquid mixtures using simple mass and volume additivity. This question tests the ability to convert a stated overall specific gravity and a known composition into the unknown specific gravity (density relative to water at the same conditions) of the second component, assuming ideal volume additivity. Such calculations appear in blending, formulation, and quality control tasks.

Given Data / Assumptions:

• Total mixture specific gravity, SG_mix = 1.000 (i.e., ρ_mix = 1.000 kg/L).
• Mass fraction of A: w_A = 0.35; mass fraction of B: w_B = 0.65.
• Specific gravity of A: SG_A = 0.70 ⇒ ρ_A = 0.70 kg/L.
• Ideal volume additivity: V_total = m_A/ρ_A + m_B/ρ_B.
• Water reference density taken as 1.000 kg/L at the stated conditions.

Concept / Approach:
Choose a convenient basis (1 kg of mixture). Convert mass fractions to masses of A and B. Use additivity of specific volumes to write the total volume. Impose the definition of mixture density (mass/volume) using the given ρ_mix to solve for the unknown ρ_B (and hence SG_B). This is a single-equation, single-unknown problem.

Step-by-Step Solution:

Verification / Alternative check:
Back-calculate: V = 0.35/0.70 + 0.65/1.30 = 0.50 + 0.50 = 1.00 L; density = 1.00 kg / 1.00 L = 1.00 kg/L, consistent with the stated mixture SG.

Why Other Options Are Wrong:

• 1.25, 1.35, 1.20: Substituting these values into the volume equation produces a total volume not equal to 1.00 L, violating the given mixture density.

Common Pitfalls:
Confusing mass fractions with volume fractions; forgetting to use specific volumes (1/ρ) additively; failing to set the mixture volume from the given density and chosen mass basis.

Final Answer:
1.30