More molarity practice1. how many grams of sodium chloride are contained in 250. ml of 0.100 m nacl?2. concentrated ammonia contains 26 g per 100 ml of solution. what is the molarity of this solution?3. what volume of 0.275m ba (oh) 2 must be used to have 8.65 g of barium hydroxide?4. what is the concentration of all ions present in a solution that is 0.250 m alcl3? 5. what mass of solute would be needed to prepare 125 ml of 0.188 m sodium phosphate?6. determine the volume of 1.55 m silver nitrate that contains 4.22 g of solute. 7. what mass of ammonium sulfate is contained in 250 ml of 0.779 m solution?8. what is the molarity of a solution that is prepared by dissolving 48.6 g of sodium carbonate in enough water to make 125 ml of solution?9. what volume of 0.88 m potassium chromate contains 0.32 moles of potassium ions?10. what volume of 0.200 m calcium nitrate contains 8.33 g of solute?

Question 1

The gram of NaCl is 1.463 g

calculation

Step 1: find moles of NaCl

moles = molarity x volume in liters

volume in liters = 250 / 1000 = 0.25 L

molarity = 0.100 M = 0.100 mol/l

moles is therefore = 0.100 mol / l x 0.25 L = 0.025 moles

Step 2; find mass of NaCl

mass = moles x molar mass

from periodic table the molar mass of NaCl=23 + 35.5 = 58.5 g/mol

mass is therefore = 0.025 moles x 58.5 g/mol = 1.463 g of NaCl

Question 2

The concentration of ammonia is 15.29 M

calculation

Step 1: find moles of NH₃

moles = mass: molar mass

from periodic table the molar mass of NH₃ = 14 + (1 x 3) = 17 g/mol

moles = 26 g:17 g/mol = 1.529 moles

Step 2: find concentration of NH₃

concentration = moles / volume in liters

volume in liters = 100/1000 = 0.1 L

concentration = 1.529 moles / 0.1 L = 15.29 M

Question 3

The volume of Ba (OH) ₂ is 0.184 L

calculation

volume = moles/molarity

moles = mass: molar mass

The molar mass of Ba (OH) ₂ = 137 + (16 + 1) ₂ = 171 g/mol

moles = 8.65 g / 171 g/mol = 0.0506 moles

volume = 0.0506/0.275 = 0.184 L

Question 4

The concentration of all ions present solution of 0.250 M AlCl₃ is

Al³⁺ = 0.250 M

Cl⁻ = 0.750 M

calculation

Step 1: write the equation for dissociation of AlCl₃

AlCl₃ → Al³⁺ + 3Cl⁻

Step 2: use the mole ratio to determine concentration of ions

AlCl₃ : Al³⁺ is 1:1 therefore the concentration of Al³⁺ is also = 0.250 M

AlCl₃ : Cl⁻ is 1:3 therefore the concentration of Cl⁻ = 0.250 x 3/1 = 0.750 M

Question 5

The mass of solute that is needed to prepare 125 ml of 0.188 M sodium phosphate = 3.854 g

calculation

Step 1: find the moles of sodium phosphate

moles = molarity x volume in liters

volume in liters = 125/1000 = 0.125 L

moles = 0.188 x 0.125 = 0.0235 moles

Step 2 : find mass of sodium phosphate

mass = moles x molar mass

from periodic table the molar mass of sodium phosphate (Na₃PO₄) = 164 g/mol

mass = 0.0235 moles x 164 g/mol = 3.854 grams

question 6

The volume of 1.55 M silver nitrate that contain 4.22 g of solute is

= 0.016 L

calculation

volume = moles/molarity

moles = mass: molar mass

from periodic table the molar mass of AgNO₃ = 107.87 + 14 + (16 x3) = 169.87 g/mol

moles = 4.22 g: 169.87 g/mol = 0.025 moles

volume is therefore = 0.025 / 1.55 = 0.016 L

Question 7

The mass of ammonium sulfate is 25.74 g

calculation

step 1: find moles of ammonium sulfate

moles = molarity x volume in liters

volume in liters = 250 / 1000=0.25 L

molarity = 0.779 M = 0.779 mol/L

moles = 0.779 mol/l x 0.25 L = 0.195 moles

Step 2: find the mass of ammonium sulfate

mass = moles x molar mass

from periodic table the molar mass of ammonium sulfate (NH₄) ₂SO4 = 132 g/mol

mass = 0.195 moles x 132 g/mol = 25.74 g

question 8

The molarity of a solution that is prepared by dissolving 48.6 g of sodium carbonate in enough water to make 125 ml of solution is 3.66 M

calculation

Step 1: find moles of sodium carbonate (Na₂CO₃)

moles = mass: molar mass

from periodic table the molar mass of Na₂CO₃ = (23 x2) + 12 + (16 x3) = 106 g/mol

moles = 48.6 g: 106 g/mol = 0.458 moles

Step 2: find the molarity

molarity = moles / volume in liters

volume in liters = 125/1000 = 0.125 L

molarity = 0.458 moles / 0.125 L = 3.66 M

question 9

The volume of 0.88 M potassium chromate that contains 0.32 moles of potassium ions is 0.18 L

calculation

Step 1: write the equation for dissociation of potassium chromate

K₂CrO₄ → 2K⁺ + CrO₄²⁻

Step 2: use the mole ratio to determine the moles of K₂CrO₄

K₂CrO₄ : K⁺ is 1:2 therefore the moles of K₂CrO₄ = 0.32 moles x1/2 = 0.16 moles

Step 3: find the volume of K₂CrO₄

volume = moles / molarity

molarity = 0.88 M = 0.88 mol/L

= 0.16 moles / 0.88 mol/L = 0.18 L

Question 10

The volume of 0.200 M calcium nitrate that contain 8.33 g of solute is 0.254 L

calculation

volume = moles / molarity

moles = mass:molar mass

from periodic table the molar mass of calcium nitrate Ca (NO₃) ₂ = 164 g/mol

moles = 8.33 g:164 g/mol = 0.0508 moles

volume = 0.0508 moles/0.200 = 0.254 L