Question

Find mass of urea ( NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.

Answer 1

Answer:

Mass of solution = 2.5 Kg

Mass of solution = 2.5 KgMolality = 0.25 m

Mass of solution = 2.5 KgMolality = 0.25 mMolar mass of urea (NH2CONH2) =(14 + 2 × 1 + 12 + 16 + 14 + 2 × 1) = 60 g mol−1

Mass of solution = 2.5 KgMolality = 0.25 mMolar mass of urea (NH2CONH2) =(14 + 2 × 1 + 12 + 16 + 14 + 2 × 1) = 60 g mol−1Mass of urea = Number of moles of urea × Molar mass of urea

Mass of solution = 2.5 KgMolality = 0.25 mMolar mass of urea (NH2CONH2) =(14 + 2 × 1 + 12 + 16 + 14 + 2 × 1) = 60 g mol−1Mass of urea = Number of moles of urea × Molar mass of ureaMass of 0.25 moles of urea = 0.25 mol × 60 g mol⁻¹ = 15 g

Mass of solution = 2.5 KgMolality = 0.25 mMolar mass of urea (NH2CONH2) =(14 + 2 × 1 + 12 + 16 + 14 + 2 × 1) = 60 g mol−1Mass of urea = Number of moles of urea × Molar mass of ureaMass of 0.25 moles of urea = 0.25 mol × 60 g mol⁻¹ = 15 gMass of solution = 1000 g + 15 g = 1015 g

Mass of solution = 2.5 KgMolality = 0.25 mMolar mass of urea (NH2CONH2) =(14 + 2 × 1 + 12 + 16 + 14 + 2 × 1) = 60 g mol−1Mass of urea = Number of moles of urea × Molar mass of ureaMass of 0.25 moles of urea = 0.25 mol × 60 g mol⁻¹ = 15 gMass of solution = 1000 g + 15 g = 1015 g1015 g of aqueous solution contains urea = 15 g

Mass of solution = 2.5 KgMolality = 0.25 mMolar mass of urea (NH2CONH2) =(14 + 2 × 1 + 12 + 16 + 14 + 2 × 1) = 60 g mol−1Mass of urea = Number of moles of urea × Molar mass of ureaMass of 0.25 moles of urea = 0.25 mol × 60 g mol⁻¹ = 15 gMass of solution = 1000 g + 15 g = 1015 g1015 g of aqueous solution contains urea = 15 g∴ 2500 g of aqueous solution will require urea = 15 g / 1015 kg × 2500 g = 36.95 g = 37 g

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Mass of urea required required in making 25 grams of 0.25m = 37 grams