An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C2 H6O2 ) and200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL-1, then what shall be the molarity of the solution?

Sure, let's solve this step by step.

Step 1: Calculate the molality of the solution.

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

First, we need to find the number of moles of ethylene glycol (C2H6O2). The molar mass of ethylene glycol is approximately 62.07 g/mol.

Number of moles = mass / molar mass Number of moles of ethylene glycol = 222.6 g / 62.07 g/mol = 3.58 mol

The mass of water (the solvent) is given as 200 g, but we need this in kilograms for the molality calculation, so we convert it to 0.2 kg.

Molality (m) = moles of solute / kg of solvent Molality (m) = 3.58 mol / 0.2 kg = 17.9 m

Step 2: Calculate the molarity of the solution.

Molarity (M) is defined as the number of moles of solute per liter of solution.

First, we need to find the volume of the solution. We know the mass of the solution (mass of solute + mass of solvent = 222.6 g + 200 g = 422.6 g) and the density of the solution (1.072 g/mL), so we can find the volume using the formula:

Volume = mass / density Volume = 422.6 g / 1.072 g/mL = 394.3 mL

We need to convert this volume to liters for the molarity calculation, so we divide by 1000 to get 0.3943 L.

Molarity (M) = moles of solute / L of solution Molarity (M) = 3.58 mol / 0.3943 L = 9.08 M

So, the molality of the solution is 17.9 m and the molarity of the solution is 9.08 M.