How many grams of glucose (C₆H₁₂O₆) must be dissolved in 750 g of ethanol (C₂H₅OH) to prepare a 0.545 m solution?
MM(C₆H₁₂O₆) = 180.15 g/mol

Question

How many grams of glucose (C₆H₁₂O₆) must be dissolved in 750 g of ethanol (C₂H₅OH) to prepare a 0.545 m solution? MM(C₆H₁₂O₆) = 180.15 g/mol

Quizplus · Accepted Answer

To find the mass of glucose needed to prepare a 0.545 m solution in 750 g of ethanol, we use the formula for molality (m), which is moles of solute per kilogram of solvent. First, calculate the moles of glucose needed: $0.545 \, \text{mol/kg} \times 0.750 \, \text{kg} = 0.40875 \, \text{mol}$. Then, find the mass of glucose: $0.40875 \, \text{mol} \times 180.15 \, \text{g/mol} = 73.6 \, \text{g}$.

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