Question

What is the `%"w/v"` of `"NaOH"` if `"4.0 g"` of it was in `"500 mL"` and `"10 mL"` of it was used to get `"100 mL"`?

Answer 1

`"0.08% w/v NaOH"`, assuming the `"500 mL"` is the final total volume of the starting solution.

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It's a little unclear exactly what you mean, but I assume you are saying:

• `"4.0 g"` of sodium hydroxide was dissolved in water and diluted to `"500 mL"`.
• `"10 mL"` of that solution was diluted to `"100 mL"`.
• We want the resultant concentration in `% "w/v"`.

We define `%"w/v"` as:

`bb(%"w/v" = "g solute"/"mL soln" xx 100%)`

We started with:

`"4.0 g NaOH"/"500 mL soln"xx100% = 0.8% "w/v NaOH"`,

because the `"mL soln"` is for the final volume (including volume changes due to its displacement by the solute).

If we then take `"10 mL"` of this solution and dilute it to `"100 mL"`, we have diluted it `10`-fold.

Since the volume multiplied by `10`, the concentration divided by `10`, as the same number of particles present in a larger volume has a smaller concentration.

We can more easily see this in the dilution equation, where `M` is concentration and `V` is volume:

`M_1V_1 = M_2V_2`

`(0.8% "w/v NaOH")("10 mL") = M_2("100 mL")`

`=> M_2 = ((0.8% "w/v NaOH")(10 cancel"mL"))/(100 cancel"mL")`

`= color(blue)(0.08% "w/v NaOH")`