how to determine molar concentration and mass What is the molar concentration of an NaOH (aq) solution that contains 12.0 grams of NaOH...

#1.  Molar concentration of an NaOH solution.  Molarity is
always based on # of moles per 1000 mL of solution.  In this case you have 12 g in 600
mL, therefore you have 20 g/L.  @0g NaOH / 40 g/mole = 0.50 M
solution.

#2.  Mass of K3PO4 required.  Molar mass is
212.27 g/mole.  For one liter of 0.2 M solution you need 212.27*.2 = 42.45 g.  For 1.5 L
you would then need 42.45 * 1.5 = 63.68 g of K3PO4

#3.  How
prepare solution of NaCl.  0.1 M means you have 0.1M x 58.5 g/mole, or 5.85 g/1000 mL. 
To make 250 mL of solution, you need 5.85/4 = 1.46 g of NaCl. Weigh out 1.46 g and
dissolve in 250 mL of water.

#4. Volume of 0.020 M KOH.  A
1.0 M solution has 56.18 g/liter. Therefore a 0.02 M solution has 56.18 * 0.02 = 1.12
g/1000 mL or 11.2 g in 10 L.

#5.  Resulting concentration. 
KBr = 119 g/mole. In 400 mL of 0.35 M you have 119 * .4 * .35 = 16.66 g.  16.66 g/500 mL
= 33.32 g/L.  33.32/119 = 0.28 M.

#6. How much water add to
make 0.75 M solution.  500/0.75 = 666.67 mL total volume. You already have 500 mL so
have to add 166.67 mL

#7. Given the equation.  molar mass
of I2 = 253.8.  Molar mass of NaI = 149.9.   508 g of I2 = 508/253.8 = ~2.0 moles. From
your equation, for every 2 moles of NaI used, you produce 1 mole of I2, therefore you
need 4 moles of NaI.  4 * 149.9 = ~600 g.  A 0.1 M solution of NaI has 15 g/L.  600/15 =
40, so you need 40 L of the 0.1 M solution.