Question

2L of 2.3%wt salt water to 5%wt salt water from how much 25%wt salt water?

A 2 L tank contains a 2.3% salt by weight. How much solution (in mL) of 25%
salt by weight would you need to add to the tank to raise the salt concentration to 5% salt by weight? The density of salt water can be computed from:
𝜌 = 998.21 + 714 𝑋𝑠
where ρ is the density measured in kg/m3 and Xs is the mass fraction of NaCl.

Answer 1

You must add 270 mL of 25 % m/m `"NaCl"` solution.

Explanation:

`"Mass of 2.3 % solution + Mass of 25 % solution = Mass of 5 % solution"`

`V_2.3ρ_2.3 + V_25ρ_25 = V_5ρ_5`

Step 1. Calculate the densities of each solution

`ρ_2.3 = "(0.998 21 + 0.714 × 0.023) g/mL" = "(0.998 21 + 0.0164) g/mL" = "1.0146 g/mL"`

`ρ_5 = "(0.998 21 + 0.714 × 0.05) g/mL" = "(0.998 21 + 0.036) g/mL" = "1.034 g/mL"`

`ρ_25 = "(0.998 21 + 0.714 × 0.25) g/mL" = "(0.998 21 + 0.178) g/mL" = "1.177 g/mL"`

Step 2. Calculate the volume of 25 % m/m `"NaCl"`

Let `V_25 = xcolor(white)(l)"mL"`.

Then

`2000 color(red)(cancel(color(black)("mL"))) × 1.0146 color(red)(cancel(color(black)("g·mL"^"-1"))) + x color(red)(cancel(color(black)("mL"))) × 1.177 color(red)(cancel(color(black)("g·mL"^"-1")))`

`= (2000 + x) color(red)(cancel(color(black)("mL"))) × 1.034color(red)(cancel(color(black)("g·mL"^"-1")))`

`2029 + 1.177x = 2068 + 1.034x`

`0.143x = 39`

`x = 39/0.143 = 270`

∴ You must add 270 mL of 25 % m/m `"NaCl"` solution.

Check:

`2000 × 1.0146 + 270 × 1.177 = 2270 × 1.034`

`2029 + 318 = 2347`

`2347 = 2347`

It checks!