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?

 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

 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

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

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

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

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

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

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

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

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

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.