Molarity Problem, calculating molarity in solution?

Question

Calculate each of the following quantities
volume of water added to 0.150 L of 0.0264 M sodium hydroxide to obtain a 0.0100 M solution (assume the volumes are additive at these low concentrations)

I have several other probelems that require conversions from/to molarity. So particular emphasis on how to use it as a conversion factor would be helpful. Thanks!

Answer 1

M=mol/L
0.0264=mol/0.150
mol=0.00396

0.0100=0.00396/L
L=0.00396/0.0100
L=0.396

So Vf-Vi=Va
Vf=final volume Vi=initial volume Va=added volume
0.396L-0.150L=0.246L


Can also do
M1V1=M2V2
0.0264M x 0.150 l= 0.0100M x V2
V2=(0.0264M x 0.150L)/0.0100M=0.396L

now solve for volume of water added
V3=V2-V1=0.396L - 0.150L= 0.246L

Answer 2

Molarity is concentration (n/v=C) first find the moles (n) by taking your mass divided over your molar mass (you will find the molar masses of Ag etc. on the periodic table of course) then you take that number and put it divided over your 0.335L of solution (because it has to be in Liters, so the final units are mol/L) Happy Chemistry-ing!

Answer 3

Use M1V1 = M2V2

(0.0264 mol per liter) (0.150 L) = (0.0100 mol per liter) (x)

The x is the total volume of the new solution, so subtract 0.150 L from it to get the volume of water that must be added.

Answer 4

The basic equation to remember is
SIGMA (volume x normality of solutions)=
volume x normaility of final solution. So for our two solutions going into the final solution
V1= 0.15
N1 = 0.0264 (normality of NaOH= molarity)
V2= ?
N2 = 0
V(final)= (V1+V2)
N(final) = 0.01
Then you get 0.01/0.0264 = 0.15/(0.15+V2)
So solve.