Question

Given `1.0*mol*L^-1` solutions of `NaCl(aq)`, what are the masses of solute in.....?

`(i)` `"a 1 L volume of solution;"`
`(ii)` `"a 250 mL volume of solution;"`
`(iii)` `"a 100 mL volume of solution?"`

Answer 1

Well the molar mass of `NaCl` is `(22.99+35.45)*g*mol^-1=` `58.44*g*mol^-1`.......

Explanation:

From where did I get the atomic masses?

For the remaining problems, we use the relationship.....

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

And so, if we gots a `1*L` volume of `1*mol*dm^-3` solution with respect there is a mass of.....

`1*Lxx1*mol*L^-1xx58,44*g*mol^-1=58.44*g`

And note that `1*dm^3-=(1xx10^-1*m)^3-=1xx10^-3*m^3=1*L`, because there are `1000*L` in a `m^3`.

And we also gots `250*cm^3xx10^-3*dm^3*cm^-3*2*mol*dm^-3xx58.44*g*mol^-1=29.22*g`

And I will let you the third.......