A 1.0L of an aqueous solution of ethylene glycol, HOCH2CH2OH, is 60% HOCH2CH2OH by mass and has a density of 1.15 g/mL. Calculate the morality of the solution.

Sorry I meant molality

molality = moles solute/Kg Solvent

It is a concentration term used mainly for systems that go through a change in temperature and need a constant value for concentration. Concentrations in terms of Molarity would change with changes in temperature due to expansion or contraction of the solution. Anyways...

moles of solute in a 60%EG(aq) soln = (60-g/62-g/mol)/dL (deciliter = 1/10th Liter = 100-ml) Soln = 0.968-mol/dL

Kg Solvent in a 60%EG(aq) soln = (100-60)gms HOH/dL Soln = 40-gms HOH/dL Soln = (40/1000)-Kg HOH/dL soln = 0.04-Kg/dL soln.

molality(m) = moles EG/Kg HOH
= (0.968 mol/dL/0.04 Kg/dL)
= 24.2 molal solution of Ethylene Glycol

To calculate the molarity of the solution, we need to know the number of moles of ethylene glycol (HOCH2CH2OH) present in the solution.

Step 1: Determine the mass of ethylene glycol in the solution
Since the solution is 60% ethylene glycol by mass, we can calculate the mass of ethylene glycol as follows:

Mass of ethylene glycol = 60% × Mass of solution
= 60% × (1.0 L) × (1.15 g/mL)
= 0.69 kg

Step 2: Convert the mass of ethylene glycol to moles
To convert the mass of ethylene glycol to moles, we need the molar mass of ethylene glycol.

Molar mass of HOCH2CH2OH = 2(1.01 g/mol) + 4(12.01 g/mol) + 2(1.01 g/mol) + 1(16.00 g/mol) + 1(1.01 g/mol) + 1(1.01 g/mol) + 1(16.00 g/mol)
= 62.07 g/mol

Now, we can calculate the number of moles of ethylene glycol:

Number of moles of ethylene glycol = Mass of ethylene glycol / Molar mass of HOCH2CH2OH
= 0.69 kg / 62.07 g/mol
= 11.13 mol

Step 3: Calculate the molarity
Finally, we can calculate the molarity of the solution using the number of moles of ethylene glycol and the volume of the solution:

Molarity of the solution = Number of moles of ethylene glycol / Volume of solution
= 11.13 mol / 1.0 L
= 11.13 M

Therefore, the molarity of the solution is 11.13 M.

To calculate the molarity of a solution, you need to know the number of moles of solute (in this case, ethylene glycol) and the volume of the solution.

First, let's calculate the mass of ethylene glycol in the solution. We know that the solution is 60% ethylene glycol by mass, so we can assume there are 60 grams of ethylene glycol in every 100 grams of solution. Since the solution has a density of 1.15 g/mL and the volume is 1.0L, the mass of the solution is 1.15 g/mL * 1000 mL = 1150 g. Therefore, the mass of ethylene glycol in the solution is 60% * 1150g = 690g.

Next, we need to convert the mass of ethylene glycol to moles. The molar mass of ethylene glycol is calculated by summing the atomic masses of its constituent elements. Carbon (C) has an atomic mass of 12.01 g/mol, Hydrogen (H) has an atomic mass of 1.01 g/mol, and Oxygen (O) has an atomic mass of 16.00 g/mol. Ethylene glycol (C2H6O2) has a molar mass of 2*12.01 + 6*1.01 + 2*16.00 = 62.07 g/mol.

So, the number of moles of ethylene glycol in the solution is 690g / 62.07 g/mol = 11.11 mol.

Now, we have all the information we need to calculate the molarity. The molarity (M) is defined as the number of moles of solute divided by the volume of the solution in liters. In this case, the volume is given as 1.0L.

Molarity = Moles of solute / Volume of solution
Molarity = 11.11 mol / 1.0 L = 11.11 M

Therefore, the molarity of the solution is 11.11 M.