Wednesday, May 18, 2011

Exam Review #13

a.       1.2 g NaCl x (1 mol NaCl/58.44 g NaCl) = .021 mol NaCl
.021/.450 = .047 M NaCl
b.      .40(.650) =.26 mol
c.       M1V1=M2V2
          .100(.200)= 12.0(V2)
          .00167 L
          1.67 mL of stock solution
           198.33 mL of water will be needed


  1. Great job kamil. My only suggestion would be to maybe add units to problem B and C. Otherwise your problem was good!

  2. Awesome Kamil! I got all the same answers. My only suggestion would to be add units like Will said just to avoid any confusion. You might also consider using math type which allows you to put subscripts on equation like for the M1V1 equation just to make it easier to read and more neat. Well done!

  3. Kamil, I agree with the value of your answers, but shech the sig figs on your pH. I get 1.000