contestada

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?

Respuesta :

kenmyna

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



Otras preguntas