An automobile antifreeze contains equal volumes of ethylene glycol (d=1.114 g/mL and FW = 62.07) and water (d=1.00 g/mL) at 25oC. The solution has a density of 1.06 g/mL. What is the molarity of ethylene glycol in the solution?
1.06 g/mL x 1000 = 1060g is mass 1000 mL soln.
This 1000 mL soln consists of 500 mL ethylene glycol and 500 mL H2O(equal volumes of ethylene glycol and H2O).
The mass of 500 mL ethylene glycol is 500 x 1.114 g/mL = 557 grams or 557/62.07 = 8.97 moles ethylene glycol per 1000 mL soln or 8.97 M. Check my thinking. I have assumed that the volume of ethylene glycol and H2O are additive which probably is not exactly true.
To find the molarity of ethylene glycol in the solution, we need to calculate the moles of ethylene glycol present and divide it by the volume of the solution in liters.
First, let's calculate the mass of the solution. We are given that the density of the solution is 1.06 g/mL, so for every 1 mL of the solution, we have 1.06 g. To find the mass of the solution, we need to multiply the volume of the solution by its density.
Given that the volume of the ethylene glycol and water mixture is equal, we can assume that they contribute equally to the total volume of the solution.
Let's assume that we have 1 L of the solution. So, the mass of the solution will be 1 L * 1.06 g/mL = 1060 g.
Now, let's calculate the mass of the ethylene glycol in the solution. Since the volume of the ethylene glycol and water mixture is equal, we have 0.5 L of ethylene glycol in the mixture. The density of ethylene glycol is 1.114 g/mL, so the mass of ethylene glycol is 0.5 L * 1.114 g/mL = 0.557 g.
Next, let's calculate the moles of ethylene glycol. To do this, we need to divide the mass of ethylene glycol by its molar mass (FW = 62.07 g/mol).
0.557 g / 62.07 g/mol = 0.008974 mol
Finally, let's calculate the molarity of ethylene glycol by dividing the moles of ethylene glycol by the volume of the solution in liters (1 L).
Molarity = moles / volume
Molarity = 0.008974 mol / 1 L
Therefore, the molarity of ethylene glycol in the solution is 0.008974 M.