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Concentration. moles solute. (. M. ). =. Molarity. liters of solution. MOLARITY A measurement of the concentration of a solution. Molarity (M) is equal to the moles of solute (n) per liter of solution M =__ n__ = mol vol L.
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moles solute ( M ) = Molarity liters of solution MOLARITY A measurement of the concentration of a solution. Molarity (M) is equal to the moles of solute (n) per liter of solution M =__n__ = mol vol L
Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution.What would be the molarity of this solution? A replacement for mol/L is often used. It is a capital M. So if you write 1.00 M for the answer, then that is correct.
Example Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in 500.0 mL of water. First calculate the moles of solute: 1.5 g NaCl (1 mole NaCl) = 0.0257 moles of NaCl 58.45 g NaCl Next convert mL to L: Last, plug the appropriate values into the correct variables in the equation: M = n / V = 0.0257 moles / 0.500 L = 0.051 mol/L 0.500 L
MOLARITY & DILUTIONM1V1 = M2V2 The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged. The moles of solute after the dilution (na) are the same: na = nb And the moles for any solution can be calculated by: n=MV A relationship can be established such that MaVa = na = nb = MbVb
Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar). M1 = 0.05 mol/L M2 = ? V1 = 25.0 mL V2 = 50.0 + 25.0 = 75.0mL M1V1 = M2V2 M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167M V2 75.0 mL of KI
MOLARITY & DILUTION Given a 6.00 M HCl solution, how would you prepare 250.0 mL of 0.150 M HCl? M1 = 6.00 mol/L M2 = 0.150 V1 = ? mL V2 = 250.0 mL M1V1 = M2V2 M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl M1 6.00 mol/L
GROUP PROBLEMS 1)10.0 g of acetic acid (CH3COOH) is dissolved in 500.0 mL of solution. What molarity results? 2) How many mL of solution will result when 15.0 g of H2SO4 is dissolved to make a 0.200 M solution? 3) Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in sea water? 4) What is the molarity of 245.0 g of H2SO4 dissolved in 1.00 L of solution? 5) What is the molarity of 5.30 g of Na2CO3 dissolved in 400.0 mL solution?
GROUP PROBLEMS 6) What is the molarity of 5.00 g of NaOH in 750.0 mL of solution? 7) How many moles of Na2CO3 are there in 10.0 L of 2.0 M solution? 8) How many moles of Na2CO3 are in 10.0 mL of a 2.0 M solution? 9) How many moles of NaCl are contained in 100.0 mL of a 0.20 M solution? 10) What weight (in grams) of NaCl would be contained in problem 9? 11) What weight (in grams) of H2SO4 would be needed to make 750.0 mL of 2.00 M solution?
12) If I add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be? 13) If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be? 14) How much 0.05 M HCl solution can be made by diluting 250 mL of 10 M HCl? 15) I have 345 mL of a 1.5 M NaCl solution. If I boil the water until the volume of the solution is 250 mL, what will the molarity of the solution be? 16) How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution?