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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.

1 Answer

6 votes

Q. No. 1 is Incomplete.

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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

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Q. No.3: A solution is prepared by dissolving 42.23 g of NHCl 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

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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

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

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