# Density of unknown solution?

A glass vessel weighs $$\pu{20.2367 g}$$ when empty and $$\pu{20.3102 g}$$ when filled to the etched mark with water at $$\pu{4 °C}$$. The same vessel was then dried and filled to the same mark with a solution at $$\pu{4 °C}$$, the vessel was now found to weigh $$\pu{20.3300 g}$$. What is the density of the solution? (Assume the density of water is $$\pu{1.00 g/ml}$$)

Here’s what I did:

The mass of the water is just $$\pu{0.0735 g}$$ when you subtract the weight of (water + vessel) and (empty vessel). Then I found how many $$\pu{ml}$$ of water there was in the vessel by realizing that the density ($$m/V$$) must be equal to $$1$$, so $$\pu{0.0735 g}$$ of water must be $$\pu{0.0735 ml}$$ of water too. So with that, I figured out how many grams of the unknown solution were in the vessel ($$\pu{0.0993 g}$$), and knowing that it was filled to the same “etched mark,” we can say that the density of the unknown solution is $$\pu{0.0933 g}/\pu{0.0735 ml} = \pu{1.27 g/ml}$$. I went through all of that because I am not sure if my thoughts are on track with logic and I am pretty confused.

What you did in your explanation is correct. First find the difference in the weight: $$20.3102-20.2367=0.0735 \ \mathrm{g}$$
We know that from your density of water: $$\mathrm{\frac{1\ g}{1\ mL}\ \text{that }\ 0.0735 \ g = 0.0735\ mL}$$
Then the difference between the unknown and the vessel is calculated as: $$20.300-20.2367 = 0.0933\ \mathrm{g}$$
and then density is expressed as mass over volume, so: $$\mathrm{\frac{0.0933\ g}{0.0735\ mL}=1.2693_8\ g\ mL^{-1}}$$