Final answer:
To calculate the amount of solute, solvent, or solution needed to produce an aqueous solution, we can use the formula: Concentration (Molarity) = moles of solute / volume of solution (in liters). We can rearrange this formula to calculate the moles of solute needed and then convert it to the mass of solute using the molar mass of the substance.
Step-by-step explanation:
Question 51:
In order to calculate the amount of solute, solvent, or solution needed to produce the given aqueous solutions, we need to use the formula:
Concentration (Molarity) = moles of solute / volume of solution (in liters)
By rearranging this formula, we can calculate the moles of solute needed and then convert it to the mass of solute using the molar mass of the substance.
Let's apply this formula to the given examples:
(a) 50.0 ml of a 0.050 M aqueous acetic acid solution CH3COOH:
To calculate the amount of acetic acid needed, we first calculate the moles of acetic acid:
Moles = Concentration x Volume
Moles = (0.050 mol/L) x (0.050 L) = 0.0025 mol
Then, we convert the moles to grams using the molar mass of CH3COOH which is 60.05 g/mol:
Mass = Moles x Molar Mass
Mass = (0.0025 mol) x (60.05 g/mol) = 0.150 g
Therefore, 0.150 g of acetic acid is needed to produce 50.0 ml of a 0.050 M aqueous acetic acid solution.
(b) 350.0 grams of a 25% w/w aqueous solution of calcium chloride CaCl2:
To calculate the amount of calcium chloride needed, we first convert the mass percentage to the mass of solute in the solution:
Mass of Solute = Mass of Solution x Mass Percentage
Mass of Solute = (350.0 g) x (25/100) = 87.5 g
Then, we convert the mass of solute to moles using the molar mass of CaCl2 which is 110.98 g/mol:
Moles = Mass of Solute / Molar Mass
Moles = (87.5 g) / (110.98 g/mol) = 0.787 mol
Therefore, 0.787 mol of calcium chloride is needed to produce 350.0 grams of a 25% w/w aqueous solution.
(c) 100.0 mL of a 0.25 M nitric acid HNO3:
To calculate the amount of nitric acid needed, we first calculate the moles of nitric acid:
Moles = Concentration x Volume
Moles = (0.25 mol/L) x (0.100 L) = 0.025 mol
Then, we convert the moles to grams using the molar mass of HNO3 which is 63.01 g/mol:
Mass = Moles x Molar Mass
Mass = (0.025 mol) x (63.01 g/mol) = 1.575 g
Therefore, 1.575 g of nitric acid is needed to produce 100.0 mL of a 0.25 M aqueous nitric acid solution.