Question

What is the concentration of a `1.40*g` mass of a `37%"(w/w)"` solution of `NaOH`? How much solute would there in `1*L` of this solution? To what volume of water would be the `NaOH` solution be added to give a concentration of `0.5*mol*L^-1`?

Answer 1

You need to quote the density of the solution.....

Explanation:

We need the quotient, `"moles of solute"/"volume of solution"`, and we cannot access this without knowledge of `rho_"density of solution"`.

This site quotes `rho_"conc. NaOH"=1.40*g*mL^-1`, and we will go from here....

And so we gots....

`((1.40*gxx37%)/(40.00*g*mol^-1))/(1.0xx10^-3*L)=12.95*mol*L^-1`

And thus per litre there are `12.95*molxx40.00*g*mol^-1=518*g` `NaOH`.

You could carefully dilute this down to a `0.5*mol*L^-1` concentration.

`C_1V_1=C_2V_2`....

`V_1=(100*mLxx0.5*mol*L^-1)/(12.95*mol*L^-1)=7.8*mL`

Add the lye to the water, and not the water to the lye.