What volume (mL) of a 5.48 M lead nitrate solution must be diluted to 391 to make a 1.84 solution of lead nitrate?

Question

What volume (mL) of a 5.48 M lead nitrate solution must be diluted to 391  to make a 1.84  solution of lead nitrate?

Answer 1

With this question you are not assigning units to the numbers 391 and 1.84. If what you mean is 391 ml and 1.84M. then you can start with showing that M = mol/L

Your substance doesn't change. You still have lead nitrate through out. So moles are the same.

So you can make mol stand for X and since substance doesn't change figure out what the volume would have to be:

5.48M=. X/0.391 L

And solve for X. Notice how we express ml into Liters for this.

X = 2.14. Now plug this number into the second equation for the second Molarity and you can find L and hence mL.

1.84M = 2.14/L.

When you solve for L you get

L = 2.14/1.84. = 1.163 L in mL it's 1160 mL for 3 sig figs.