Question

What mass of solute is present in a `0.250*L` volume of `NaOH` solution, for which `[NaOH]-=2.50*mol*L^-1`?

Answer 1

Approx. `25*g.................`

Explanation:

First we calculate the molar quantity present in the given volume of `NaOH`........

`"Molarity"="Moles of solute"/"Volume of solution"...........`

And thus `"Moles of solute"="Volume"xx"molarity"`.....

`250*xx10^-3*cancelLxx2.5*mol*cancel(L^-1)=0.625*mol`, with respect to `NaOH`...............

which molar quantity corresponds to a MASS of..........

`0.625*molxx40.0*g*mol^-1=??*g`

You must simply KNOW that `1*mL-=1xx10^-3*L`, i.e. the `"milli"` prefix `-=10^-3`.