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