kyliejones2002
contestada

1). If the molarity of a KCl (potassium chloride) solution is 8.0 M, which of these is true?
2.) A saline solution contains 0.015 mol NaCl in exactly 0.10 L of solution. What is the molarity of the solution?
3.) A solution is prepared by dissolving 42.23 g of NH4Cl into enough water to make 0.500 L of solution. Calculate its molarity.
4.)Determine the concentration of a solution made by dissolving 10.0 g of sodium chloride (NaCl) in 0.75 L of solution.
5.)Determine the concentration of a solution made by dissolving 44.0 g of calcium chloride (CaCl2) in 0.30 L of solution.

Respuesta :

transitionstate

Q. No. 1 is Incomplete.

___________________________________________

Q. No.2: A saline solution contains 0.015 mol NaCl in exactly 0.10 L of solution. What is the molarity of the solution?

Answer:

                  Molarity  =  0.15 mol.L⁻¹ (or) 0.15 M

Solution:

                    Molarity is the the unit of concentration and it is expressed as the amount of solute dissolved per unit volume of solution. It is expressed as,

                        Molarity  =  Moles / Volume of Solution  (1)

Data Given;

                        Moles  =  0.015 mol

                        Volume  =  0.10 L

Now, putting value of Moles and Volume in eq. 1,

                        Molarity  =  0.015 mol ÷ 0.10 L

                        Molarity  =  0.15 mol.L⁻¹ (or) 0.15 M

___________________________________________

Q. No.3: A solution is prepared by dissolving 42.23 g of NH₄Cl into enough water to make 0.500 L of solution. Calculate its molarity.

Answer:

                  Molarity  =  1.578 mol.L⁻¹ (or) 1.578 M

Solution:

Data Given;

                 Mass  =  42.23 g

                 Volume  =  0.50 L

                 M.Mass of NH₄Cl  =  53.49 g/mol

First calculate Moles for given mass as,

                  Moles  =  Mass / M.mass

                  Moles  =  42.23 g / 53.49 g.mol⁻¹

                  Moles  =  0.789 mol

Formula used,

                   Molarity  =  Moles / Vol. of Solution  ---- (1)

Now, putting value of Moles and Volume in eq. 1,

                   Molarity  =  0.789 mol ÷ 0.50 L

                   Molarity  =  1.578 mol.L⁻¹ (or) 1.578 M

___________________________________________

Q. No.4: Determine the concentration of a solution made by dissolving 10.0 g of sodium chloride (NaCl) in 0.75 L of solution.

Answer:

                  Molarity  =  0.228 mol.L⁻¹ (or) 0.228 M

Solution:

Data Given;

                 Mass  =  10.0 g

                 Volume  =  0.75 L

                 M.Mass of NaCl  =  58.44 g/mol

First calculate Moles for given mass as,

                  Moles  =  Mass / M.mass

                  Moles  =  10.0 g / 58.44 g.mol⁻¹

                  Moles  =  0.171 mol

Formula used,

                   Molarity  =  Moles / Vol. of Solution  ---- (1)

Now, putting value of Moles and Volume in eq. 1,

                   Molarity  =  0.171 mol ÷ 0.75 L

                   Molarity  =  0.228 mol.L⁻¹ (or) 0.228 M

___________________________________________

Q. No.5: Determine the concentration of a solution made by dissolving 44.0 g of calcium chloride (CaCl₂) in 0.30 L of solution.

Answer:

                  Molarity  =  1.32 mol.L⁻¹ (or) 1.32 M

Solution:

Data Given;

                 Mass  =  44.0 g

                 Volume  =  0.30 L

                 M.Mass of CaCl₂  =  110.98 g/mol

First calculate Moles for given mass as,

                  Moles  =  Mass / M.mass

                  Moles  =  44.0 g / 110.98 g.mol⁻¹

                  Moles  =  0.396 mol

Formula used,

                   Molarity  =  Moles / Vol. of Solution  ---- (1)

Now, putting value of Moles and Volume in eq. 1,

                   Molarity  =  0.396 mol ÷ 0.30 L

                   Molarity  =  1.32 mol.L⁻¹ (or) 1.32 M

Otras preguntas