A test solution of sodium acetate is to contain 10 mg/ml of the metal ion. if 750 ml of the solution is needed, how many grams of sodium acetate must be used? What is the molarity of this solution?
To determine the number of grams of sodium acetate required, we first need to calculate the total amount of sodium acetate needed in the solution.
Given:
Desired concentration = 10 mg/ml
Volume of solution needed = 750 ml
To calculate the mass of sodium acetate, we can use the equation:
Mass = Concentration x Volume
Mass = 10 mg/ml x 750 ml = 7500 mg
Since 1 g = 1000 mg, we can convert the mass from milligrams to grams:
Mass = 7500 mg / 1000 = 7.5 g
Therefore, 7.5 grams of sodium acetate must be used to prepare the solution.
Now, let's determine the molarity of the solution.
Molarity is defined as the number of moles of solute per liter of solution. To calculate the molarity, we need to know the molecular weight of sodium acetate. The molecular formula for sodium acetate is NaC2H3O2.
Molar mass of sodium acetate (NaC2H3O2):
Na = 23 g/mol
C = 12 g/mol
H = 1 g/mol (x2)
O = 16 g/mol (x2)
Molar mass of sodium acetate = 23 + (12 x 2) + (1 x 2) + (16 x 2) = 82 g/mol
To calculate the molarity, we can use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, let's calculate the number of moles of sodium acetate:
Moles = Mass / Molar mass
Moles = 7.5 g / 82 g/mol = 0.091 moles
Next, we need to convert the volume of the solution from milliliters to liters:
Volume = 750 ml / 1000 = 0.75 L
Now we can calculate the molarity:
Molarity = 0.091 moles / 0.75 L = 0.121 M
Therefore, the molarity of the sodium acetate solution is 0.121 M.