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 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
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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 (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
