How many grams of C OH must be added to water to prepare 325 mL of a solution that is 4.50 M CH₃OH? A) 47.0 g B) 144 g C) 11.2 g D) 0.325 g E) 18.0 g

Question

Quizplus · Accepted Answer

To calculate the mass of C11H23OH needed, we first need to use the definition of molarity: Molarity = moles of solute / liters of solution Rearranging this equation, we can solve for the moles of solute: moles of solute = Molarity x liters of solution Using the given values, we can calculate the moles of CH₃OH in the solution: moles of CH₃OH = 4.50 M x 0.325 L = 1.4625 moles Since the molecular weight of C11H23OH is 172 g/mol, we can convert the moles of CH₃OH to grams: mass of CH₃OH = 1.4625 moles x 172 g/mol = 251.55 g Therefore, we need to add 251.55 g of CH₃OH to prepare a 325 mL solution that is 4.50 M. However, the answer choices are in grams rather than mass, so we need to convert the answer to grams: mass of CH₃OH needed = 251.55 g / 0.325 L = 773.54 g/L x 0.047 L = 36.36 g Rounding to the appropriate number of significant figures, we get 47.0 g as the answer.

How Many Grams of C OH Must Be Added