Please help me solve this questions for solution concentratrions. I don\'t reall
ID: 889766 • Letter: P
Question
Please help me solve this questions for solution concentratrions. I don't really know how to get to the answer.
1) Calculate the molarity of a solution made by dissolving 123.4 g of sodium sulfate (Na2SO4) in enough water to form 525.0mL of solution.
2) What is the molarity of a solution that is made by dissolving 13.68g of sucrose (C12H22O11) in sufficient water to form 225.0mL of solution?
3) Calculate the molarity of a solution made by dissolving 15.00 g of glucose (C6H12O6) in sufficient water to form exactly 200mL of solution.
4) What is the molar concentration of each ion present in a 0.225 M aqueous solution of calcium nitrate?
5) What is the ratio of the concentration of potassium ions to the concentration of carbonate ions in a 0.1275 M solution of Aluminium carbonate?
6) What is the molar concentration of K+ ions in a 0.0175 M solution of potassium Phosphate?
7) How many grams of Na2SO4 are required to make 1.250 L of 0.800 M Na2SO4?
8) What is the concentration of ammonia in a solution made by dissolving 13.75 g of ammonia in 8.0 L of water?
9) (a) How many grams of Na2SO4 are there in 250 mL of 0.50M Na2SO4? (b) How many milliliters of 0.50 M Na2SO4 solution are needed to provide 0.038 mol of this salt? (c)How many milliliters of 2.5M H2SO4 are needed to make 450mL of 0.20M H2SO4?
10) What volume of 1.00M stock solution of glucose must be used to make 200.0mL of a 2.75x10–2M glucose solution in water?
11) (a) What volume of 2.80 M lead(II) nitrate solution contains 0.550 mol of Pb2+? (b) How many milliliters of 5.0 M K2Cr2O7 solution must be diluted to prepare 550 mL of 0.20 M solution? (c) If 20.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting stock solution?
12) Glacial acetic acid with density 1.049 g/ml – Calculate the molarity if 20.00 ml of glacial acetic acid dissolved in enough water to produce 250.00 ml of the solution.
13) Glycerol, C3H8O3, with density of 1.2656 g/ml. Calculate the molarity of a solution of glycerol, when 50.00 ml of glycerol dissolved in enough water to make 500.00 ml of the solution.
Explanation / Answer
1). Molarity of 123.4 g Na2SO4
Given : Mass of Na2SO4 = 123.4 g
Volume of solution = 525.0 mL
We know molarity = Moles of solute / volume of solution in L
Calculation of moles of Na2SO4 = Mass of Na2SO4 in g / Molar mass of Na2SO4
= 123.4 g / 142.042 g per mol
= 0.8688 mol
Volume in L = 525.0 mL x 1 L / 1000 mL
= 0.525 L
[Na2SO4]= 0.8688 mol / 0.525 L
= 1.655 M
2).
Molarity of sucrose solution
Calculation of moles of Sucrose solution = 13.68 g/ 342.2948 g per mol
= 0.03997 mol
[Sucrose] = 0.03997 mol / 0.225 L
= 0.00899 M
3)
Molarity of glucose
Moles of glucose = 15.00 g / 180.1548 g per mol
= 0.083262 mol
Molarity = 0.083262 mol / 0.200 L = 0.4163 M
4) concentration of each in Calcium nitrate
1 mol Ca(NO3)2 contains
1 mol Ca2+ and 2 mol NO3- ions.
Calculation of moles of Ca(NO3)2
Lets assume solution is 1.L
Molarity = Number of moles
Therfore number of moles of Ca(NO3)2 = 0.225
Number of moles of Ca2+ = Number of moles of Ca(NO3)2
= 0.225 mol
Therefore [Ca2+] = 0.225 M
Number of moles of NO3- = 2 x moles of Ca(NO3)2
= 2 x 0.225 mol
= 0.45 mol NO3-
[NO3-]= 0.45 M