Ethylene glycol is the primary component in antifreeze. How many grams of ethylene glycol, C₂H₄(OH)₂ are in 394 mL of a 9.087 M solution of antifreeze? (molar mass C₂H₄(OH)₂ = 62 g\mol)

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Quizplus · Accepted Answer

To find the grams of ethylene glycol, use the formula: grams = molarity × volume (L) × molar mass. Convert 394 mL to liters (0.394 L), then multiply by the molarity (9.087 M) and the molar mass (62 g/mol): 0.394 L × 9.087 mol/L × 62 g/mol = 222 g.

Ethylene Glycol Is the Primary Component in Antifreeze