Calculations involving Molarity?

Question

Calculate each of the following quantities.
(a) volume of 14.7 M sulfuric acid that must be added to water to prepare 2.00 L of a 0.271 M solution

(b) molarity of the solution obtained by diluting 75.7 mL of 0.231 M ammonium chloride to 0.250 L

(c) volume of water added to 0.150 L of 0.0252 M sodium hydroxide to obtain a 0.0100 M solution (assume the volumes are additive at these low concentrations)

(d) mass of calcium nitrate in each milliliter of a solution prepared by diluting 56.5 mL of 0.726 M calcium nitrate to a final volume of 0.100 L

Answer 1

The no of moles of sulfuric acid in 2.00L of 0.271M solution is 2*0.271 = 0.542 moles. That no of moles is contained in 0.542/14.7 L of 14.7 M solution, or 0.0369 L

75.7mL of 0.231M ammonium chloride has
0.0757L * 0.231 moles/L = 0.0175 moles. This is now in 0.250L, so the molarity is 0.0175/0.250 = 0.07M


The 0.150 L of 0.0252M solution contains 0.150*0.0252 moles of sodium hydroxide = 0.00378 moles. New volume must be .00378/0.01 = 0.378L. So you must add 0.378 - 0.150 = .228 L

The molarity after dilution is 0.056.5*0.726/0.100 = 0.041M
One L has 0,041 moles of Calcium Nitrate, one mL has 0.041*10^-3 moles. Multiply that time the molecular mass of Calcium nitrate (164) to get 0.00673 grams

Answer 2

ok molarity is equivalent mols in keeping with liter. So because of the fact cm^3 are equivalent to mL you may convert dm^3 to cm^3 and then to mL and ultimately L and additionally you will possibly need your concentration via weight. Then multiply the consequent weight of H2SO4 via 0.02 and additionally you gets the molarity of your needed answer.

Answer 3

I think that all of these can be solved by using the basic formula M1V1 = M2V2. You have 3 values for each so plug in and solve to find the fourth. Remember to convert mL to L for the ones that need it.

For example a) would be (using M1V1=M2V2):

(0.271)(2.00)=(14.7)(x)
x=((0.271)(2.00))/(14.7)
x would be your new volume.

hope that helps.