Question

What main steps should be carried out to prepare 250 mL of 0.25 `mol L^(-1)` lead (II) nitrate, `Pb(NO_3)_2` solution?

The answer must be based on calculating the mass of `Pb(NO_3)_2` required to prepare the solution.

I got 20.7 grams as the calculated mass.

Answer 1

Well, what is the molar mass of lead nitrate?

Explanation:

I make `Pb(NO_3)_2` to be `331.20* g*mol^-1`.

Now by definition, `"molarity"-="moles of solute"/"volume of solution"`..

And thus... `"moles of solute"="molarity"xx"volume of solution"`

`-=0.250*mol*L^-1xx250*mLxx1xx10^-3*L*mL^-1=0.06250*mol`

And this represents a mass of...`0.06250*molxx331.20*g*mol^-1=??*g...`

And so I don't think you are far out....

And this mass, is transferred quantitatively to a `250*mL` volumetric flask....the flask is then diluted with a little distilled water, and the solute bought into solution.. The flask is then made up to volume...and used as required...