The density of a 22.0% by mass ethylene glycol (C2 H6 O2) solution in water is 1.03 .Find the molarity of the solution.
To find the molarity of the ethylene glycol solution, we need to understand the relationship between mass percent, molarity, and density.
First, let's determine the mass of ethylene glycol (C2H6O2) in the solution. We know that the solution is 22.0% by mass, which implies that 22.0 grams of ethylene glycol are present per 100 grams of solution.
Next, we can calculate the mass of the solution. Given the density of the solution is 1.03 g/mL, we know that 1 mL of the solution weighs 1.03 grams.
Since we don't know the total volume of the solution, we can assume a reasonable volume, such as 100 mL (which is equivalent to 100 grams).
Using these values, we can calculate the mass of ethylene glycol in the solution:
Mass of ethylene glycol = (22.0/100) * mass of solution
= (22.0/100) * 100 grams
= 22.0 grams
Next, we can find the molar mass of ethylene glycol (C2H6O2):
Molar mass of ethylene glycol = (2 * atomic mass of carbon) + (6 * atomic mass of hydrogen) + (2 * atomic mass of oxygen)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (2 * 16.00 g/mol)
= 62.07 g/mol
To calculate the molarity, we need to find the number of moles of ethylene glycol in the solution. We can use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
Now, let's convert the mass of ethylene glycol to moles:
Moles of ethylene glycol = mass of ethylene glycol / molar mass of ethylene glycol
= 22.0 g / 62.07 g/mol
≈ 0.354 mol
Since the volume of the solution is 100 mL (or 0.1 L), we can calculate the molarity:
Molarity = moles of solute / volume of solution
= 0.354 mol / 0.1 L
= 3.54 M
Therefore, the molarity of the 22.0% by mass ethylene glycol solution is approximately 3.54 M.