Ethylene glycol is the primary component in antifreeze. How many grams of ethylene glycol (C ₂ H ₄ (OH) ₂ ) are present in 394 mL of a 9.087 M solution of antifreeze (molar mass C ₂ H ₄ (OH) ₂ = 62 g/mol)? A) 222 g B) 43.7 g C) 3.58 g D) 2.69 g E) 0.157 g

Question

Quizplus · Accepted Answer

First, we need to use the molarity equation to find the number of moles of ethylene glycol present in the solution:

Molarity = moles of solute / liters of solution

Rearranging the equation gives:

moles of solute = Molarity x liters of solution

Converting the volume of the solution from milliliters to liters, we have:

394 mL = 0.394 L

Substituting the values we have:

moles of solute = 9.087 mol/L x 0.394 L = 3.582 g

Next, we need to convert moles to grams using the molar mass of ethylene glycol:

3.582 mol x 62 g/mol = 221.84 g

Rounding to the nearest tenth gives the final answer of:

222 g

Ethylene Glycol Is the Primary Component in Antifreeze