Question

What is the molarity of a 50% rubidium hydroxide solution `"(w/w)"` whose density is `1.74*g*mL^-1`?

Answer 1

Now we are given that `"Mass of solute"/"Mass of solution"xx100%=50%`.

Explanation:

We are further given `rho_"RbOH solution"=1.74*g*mL^-1`.

We know that the `"molar mass"` of `RbOH=102.48*g*mol^-1`. (And, in fact, as I recall, you can only buy this as an aqueous solution.)

And we can work out the molar concentration on the basis of these data, given a `1*mL` volume of `RbOH(aq)`

`"Concentration"` `=` `"Moles of RbOH"/"Volume of solution (L)"`

`=((1*cancel(mL)xx1.74*cancelg*cancel(mL^-1)xx50%)/(102.48*cancelg*mol^-1))/(1xx10^-3*L)`

`=8.50*mol*L^-1`

This is dimensionally consistent; the calculation gave units of `1/(mol^-1)xxL^-1=1/(1/(mol))xxL^-1=mol*L^-1` as required.