Question

What mass of solute is present in `24*mL` of `0.10*mol*l^-1` aqueous sodium chloride solution?

Answer 1

Approx. `0.140*g`.

Explanation:

We have `0.1*mol*dm^-3` `NaCl`.

Now believe or not, `1*dm^3=1*L`, because the prefix `"deci"-=10^-1`

And so,...........`1*dm^3=(1xx10^-1*m)^3`

`=1xx10^-3*m^3-=1*L`, because there are `1000*L` in a `m^3`. A `"cubic meter"` is A HUGE VOLUME..........

And so we simply perform the product:

`"Mass of NaCl"="Concentration"xx"Volume"xx"molar mass, NaCl"`

`0.024*Lxx0.1*mol*L^-1xx58.44*g*mol^-1=0.140*g`.