Question

Question #a1ef4

Answer 1

`"31.3 g"`

Explanation:

Start by making sure that you have clear understanding of what a `"25%w/v"` sodium chloride solution looks like.

A weight by volume solution is defined as mass of solute, which in your case is sodium chloride, divided by the volume of the solution, and multiplied by `100`.

`color(blue)(%"w/v" = "mass of solute"/"volume of solution" xx 100)`

This means that a `25%"w/v"` sodium chloride solution would contain `"25 g"` of sodium chloride for every `"100 mL"` of solution.

You know that you dissolve the sodium chloride in `"125 mL"` of water, so that will be your final solution's volume.

If you know the target weight by volume concentration, and what the volume of the solution is, you can solve the above equation for the mass of sodium chloride

`m_"solute" = (%"w/v" xx V_"solution")/100`

`m_"solute" = (25"g"/color(red)(cancel(color(black)("mL"))) * 125color(red)(cancel(color(black)("mL"))))/100 = "31.25 g"`

I'll leave this rounded off to three sig figs, just for good measure

`m_"NaCl" = color(green)("31.3 g")`