In a similar experiment, a sample of tin metal was dissolved in nitric acid to produce 23.00 mL of solution containing Sn2+. Titration analysis resulted in the calculation that the concentration of Sn2+ in this solution was 0.65 mol/L.
b. What mass of tin was used to make the tin solution?
Since Sn ==> Sn^2+, then if M Sn^2+ is 0.65, so mols Sn^2+ = M x L = 0.65 x 0.023 = ? and same mols Sn to have started the process.
Then g Sn = mols Sn x atomic mass Sn = ?
To determine the mass of tin used to make the tin solution, we need to use the given information about the concentration of Sn2+ and the volume of the solution.
Step 1: Convert the volume of the solution from milliliters (mL) to liters (L).
Given: Volume of solution = 23.00 mL
1 L = 1000 mL, so the volume of solution is 23.00 mL ÷ 1000 mL/L = 0.023 L.
Step 2: Use the formula for molarity to calculate the number of moles of Sn2+ in the solution.
Given: Concentration of Sn2+ = 0.65 mol/L
Number of moles of Sn2+ = Concentration × Volume = 0.65 mol/L × 0.023 L = 0.01495 mol.
Step 3: Determine the molar mass of tin (Sn).
The molar mass of tin is found on the periodic table. Sn has a molar mass of 118.71 g/mol.
Step 4: Use the molar mass of tin to convert moles of Sn2+ to grams of tin.
Moles of Sn × Molar mass of Sn = Grams of Sn
0.01495 mol × 118.71 g/mol = 1.773 g
Therefore, the mass of tin used to make the tin solution is 1.773 grams.