What is the molality of a solution prepared by dissolving 13.2 g of MgCl_2 in 175.0 g of H_2O?

• A. 0.78 mol/kg
• B. 1.20 mol/kg
• C. 0.50 mol/kg
• D. 1.00 mol/kg

Answer: (A)

Verified step by step guidance

1. Identify the formula for molality, which is defined as the number of moles of solute per kilogram of solvent: \(molality = \frac{moles\ of\ solute}{kilograms\ of\ solvent}\).
2. Calculate the moles of MgCl\(_2\) by using its molar mass. First, find the molar mass by adding the atomic masses: Mg (24.31 g/mol) + 2 × Cl (2 × 35.45 g/mol). Then, calculate moles using \(moles = \frac{mass}{molar\ mass}\).
3. Convert the mass of the solvent (water) from grams to kilograms by dividing by 1000: \(mass\ of\ solvent\ (kg) = \frac{175.0\ g}{1000}\).
4. Substitute the calculated moles of MgCl\(_2\) and the mass of water in kilograms into the molality formula: \(molality = \frac{moles\ of\ MgCl_2}{mass\ of\ H_2O\ (kg)}\).
5. Perform the division to find the molality of the solution, which will give the concentration in mol/kg.