# 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"#?

##### 1 Answer

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

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

#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 **final volume** (including volume changes due to its displacement by the solute).

If we then take

Since the volume *multiplied* by *divided* by

We can more easily see this in the **dilution equation**, where

#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")#