How Do You Calculate the Specific Gravity of Alcohol?

[Soojin]

Homework Statement

An aluminum cylinder weighs 1.03 N. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is 3.89 multiplied by 10-5 m3. If the cylinder is suspended from a scale while submerged in the alcohol, the scale reading is 0.732 N. What is the specific gravity of the alcohol?

Homework Equations

F_scale + F_Buoyancy = mg

F_Bouyancy = DVg

Specific gravity = D_object / D_medium

The Attempt at a Solution

First I used F_scale + F_Buoyancy = mg
(.732)(F_B) = 1.03
F_B = .298 N

Then I used F_B = DVg
.298 = D_alcohol(3.89e-5)(9.8)
D_alcohol = .298 / [(3.89e-5)(9.8)]
D_alcohol = 781.7009 kg.m³

Then I used Specific gravity = D_object / D_medium, where D = mv.

Specific gravity = (m/v) / D_alcohol
Specific gravity = [(1.02/9.8)/3.85e-5] / (781.7009)
Specific gravity = 3.458

I tried this answer and it was wrong. Can anyone help out? Thank you!

 

[Doc Al]

Then I used Specific gravity = D_object / D_medium

The specific gravity of a substance is the ratio of the density of that substance to the density of water.