# How do you find the mass of a solution if you know the volume and identity of the solute added, the density of the solvent, and the concentration of the solution?

##### 2 Answers

You're leaving us a bit in the dark with this question.......

#### Explanation:

If

Note that this is a measurement SOLELY of

You have not quoted units of concentration, i.e.

I think it's possible to find the density of the aqueous solution without knowing its total mass, but I think we need the volume of the solution in addition to the volume of the solute. With this information we can find the total mass of the solution by knowing the density of water at given temperature.

Suppose we are at

#V_"solvent" = V_"solution" - V_"solute"#

We can assume we have

#m_"solvent" = rho_"solvent"V_"solvent"#

And we can use the molar mass of the solute to get its mass. Therefore:

#"mol solute"/"L solution" = "mol solute"/"L solute + L solvent"#

#= (("mol/g solute") xx ("g solute"))/("L solute" + ("g solvent")/"g/L solvent"#

Once we have the mass of the solvent and solute, we can add them together (and masses are surely additive always) to get the mass of the solution.

Therefore, knowing the mass and volume of the solution, we can get its density.

**EXAMPLE**

Let's say we are given that a certain mass percent solution of

It's probably not a useful calculation to do, but let's see how this works out because this is something we can easily look up the answer to.

#V_"water" = "1.00 L soln" - "0.25 L solute"#

#~~# #"0.75 L"#

#m_"water" = "998.2071 g/L water" xx "0.75 L water"#

#=# #"748.6553 g water"#

From the concentration, we'd find the mass of the solute:

#(12.06 cancel"mol HCl")/cancel"L soln" xx "36.4609 g HCl"/cancel"mol HCl" xx cancel"1 L soln"#

#=# #"439.7185 g HCl"#

And from these masses we'd get the mass of the solution:

#"439.7185 g HCl"# #+# #"748.6553 g water"#

#=# #"1188.3738 g solution"#

Knowing its assumed volume then, we'd have:

#"1188.3738 g soln"/cancel"1 L soln" xx cancel"1 L"/"1000 mL soln"#

#~~# #color(blue)("1.19 g/mL soln")#

So I think we can do it, but we would still need the total volume of the solution.