Final answer:
To find the grams of solute in a solution, multiply the volume by the molarity and the molar mass of the solute. For Part A, the answer is 1342.1 grams of Al(NO₃)₃.
Step-by-step explanation:
To determine the number of grams of solute in a solution, you need to multiply the volume of the solution (in liters) by the molarity of the solution (in moles per liter) and the molar mass of the solute (in grams per mole).
For part A, the volume is 4.5 L and the molarity is 1.4 M. The molar mass of Al(NO₃)₃ is 213.0 g/mol. So:
Moles of Al(NO₃)₃ = volume x molarity = 4.5 L x 1.4 M = 6.3 moles
Grams of Al(NO₃)₃ = moles x molar mass = 6.3 moles x 213.0 g/mol = 1342.1 grams
To calculate the grams of solute in a solution given its molarity, you need to use the formula:
mass (g) = molarity (M) × volume (L) × molar mass (g/mol)
For Part A, we have a 1.4 M solution of Al(NO₃)₃ and a volume of 4.5 L. First, we need to find the molar mass of Al(NO₃)₃ by adding the atomic masses of Al (26.98 g/mol) and 3 × (N (14.01 g/mol) + 3 × O (16.00 g/mol)). This gives us a molar mass of 213.00 g/mol. Then, we use the formula:
mass = 1.4 M × 4.5 L × 213.00 g/mol
Calculating this, we get:
mass = 1.4 × 4.5 × 213.00 = 1335.3 grams of Al(NO₃)₃