Final answer:
To prepare a 125-mL, 0.159 M solution of chromium(III) nitrate, a student needs to weigh out 7.95 grams of the solid compound, using its molar mass to convert from moles required to grams.
Step-by-step explanation:
To calculate the amount of solid chromium(III) nitrate needed to prepare a 0.159 M solution in a 125-mL flask, we use the formula for molarity (M), which is moles of solute per liter of solution. First, we determine the number of moles needed:
- Molarity (M) = 0.159 mol/L
- Volume (V) = 125 mL = 0.125 L (since 1 L = 1000 mL)
The number of moles needed is given by:
n = Molarity (M) × Volume (V) = 0.159 mol/L × 0.125 L = 0.019875 moles
To find the mass in grams, we multiply the number of moles by the molar mass of chromium(III) nitrate, Cr(NO₃)₃. The molar mass can be calculated by summing the atomic masses of chromium (52.00 g/mol), nitrogen (14.01 g/mol), and oxygen (16.00 g/mol) from the periodic table:
Molar mass of Cr(NO₃)₃ = Cr + 3×(N + 3×O)
Molar mass of Cr(NO₃)₃ = 52.00 g/mol + 3×(14.01 g/mol + 3× 16.00 g/mol) = 400.04 g/mol
Mass of Cr(NO₃)₃ = Number of moles × Molar mass = 0.019875 moles × 400.04 g/mol = 7.95 g
The student must weigh out 7.95 g of solid chromium(III) nitrate to prepare the desired solution.