Question

2. A glass stopper weights 2.5 gm in air, and 1.5 gm in water and 0.7 gm in sulphuric acid. What is the density of the acid?

Answer 1

Let's start with specific gravity: 

"Specific gravity is the ratio of the density of a substance compared to the density (mass of the same unit volume) of a reference substance. . . . The reference substance is nearly always water for liquids . . . " (see Source 1) 

In equation form: 

(1) SGacid = ρacid / ρwater 

This is equivalent to: 

(2) SGacid = Wvacid / Wvwater (see Source 2), 

where: 
Wvacid = weight of a volume of acid 
Wvwater = weight of an equal volume of water 

Let's use the "glass stopper" as the unit of volume. Since the stopper weighs 0.7g in the acid and 2.5g in air, a stopper-volume of acid provides a buoyancy of: 

(3) Fb = 2.5 - 0.7 = 1.8g (force-weight); 

so that our unit volume of acid weighs 1.8g. 

By the same logic, the water weight is given by: 

(4) Fb = 2.5 - 1.5 = 1.0g because it provides that much buoyancy on the stopper; 

so that our unit volume of water weighs 1.0g 

Substituting (3) and (4) into (2): 

(5) SGacid = 1.8g / 1.0 = 1.8 


Now we can go back to (1), given our SGacid: 

(6) 1.8 = ρacid / ρwater 

Solving for ρacid and substituting ρwater = 1.0g/cm^3 we get: 

(7) ρacid = 1.8 / 1.0 = 1.8g/cm^3 

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