Question

molarity practice worksheet
molarity = \frac{moles solute}{liters solution} \quad m = \frac{moles}{l}

1. what is the molarity of a solution made with 298g of kcl dissolved in enough water to make 2l of solution?
2. what is the molarity of a solution made with 87g of mgo dissolved in enough water to make 800ml of solution?
3. what is the molarity of a solution prepared from 100g of alf₃ dissolved in 1200ml of water?
4. what mass of nano₃ is needed to make 400ml of a 0.3m solution?
5. what mass of sucrose (c₁₂h₂₂o₁₁) is required to prepare 4.6l of a 0.4m solution?
6. what mass of ca(oh)₂ is needed to make 1250ml of a 0.75m solution?

Explanation:

Problem 1:

Step1: Calculate moles of KCl. Molar mass of KCl: K (39.10 g/mol) + Cl (35.45 g/mol) = 74.55 g/mol. Moles = mass / molar mass = 298 g / 74.55 g/mol ≈ 3.997 mol.

$n = \frac{298\ g}{74.55\ g/mol} \approx 3.997\ mol$

Step2: Use molarity formula. M = moles / liters. Volume is 2 L. M = 3.997 mol / 2 L ≈ 2.0 M.

$M = \frac{3.997\ mol}{2\ L} \approx 2.0\ M$

Step1: Molar mass of MgO: Mg (24.31 g/mol) + O (16.00 g/mol) = 40.31 g/mol. Moles = 87 g / 40.31 g/mol ≈ 2.158 mol.

$n = \frac{87\ g}{40.31\ g/mol} \approx 2.158\ mol$

Step2: Convert volume to liters: 800 mL = 0.8 L. M = 2.158 mol / 0.8 L ≈ 2.70 M.

$M = \frac{2.158\ mol}{0.8\ L} \approx 2.70\ M$

Step1: Molar mass of AlF₃: Al (26.98 g/mol) + 3×F (19.00 g/mol) = 26.98 + 57.00 = 83.98 g/mol. Moles = 100 g / 83.98 g/mol ≈ 1.191 mol.

$n = \frac{100\ g}{83.98\ g/mol} \approx 1.191\ mol$

Step2: Convert volume to liters: 1200 mL = 1.2 L. M = 1.191 mol / 1.2 L ≈ 0.993 M.

$M = \frac{1.191\ mol}{1.2\ L} \approx 0.993\ M$

Answer:

Approximately 2.0 M

Problem 2: