Question

calculate the mass of urea required in making 2.5kg of 0.25 molar aqueous solution.​

Answer 1

$\huge\mathfrak{Answer}$

Mass of solution = 2.5 Kg

Number of moles of urea = 0.25 m

Molar mass of urea (NH2CONH2)

Molar mass of urea

=(14 + 2 × 1 + 12 + 16 + 14 + 2 ×1)

Molar mass of urea = 60 g mol−1

Mass of urea = Number of moles of urea × Molar mass of urea

Mass of 0.25 moles of urea = 0.25 mol × 60gmol⁻¹

Mass of 0.25 moles of urea = 15 g

Mass of solution = 1000 g + 15 g = 1015 g

1015 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

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

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Answer 2

Answer:

Mass of solution = 2.5 Kg

Number of moles of urea = 0.25 m

Molar mass of urea (NH2CONH2)

Molar mass of urea=(14 + 2 × 1 + 12 + 16 + 14 + 2 ×1)

Molar mass of urea = 60 g mol−1

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

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

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

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

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

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

Mass of solution = 2.5 KgNumber of moles of urea = 0.25 mMolar mass of urea (NH2CONH2)Molar mass of urea=(14 + 2 × 1 + 12 + 16 + 14 + 2 ×1)Molar mass of urea = 60 g mol−1Mass of urea = Number of moles of urea × Molar mass of ureaMass of 0.25 moles of urea = 0.25 mol × 60gmol⁻¹Mass of 0.25 moles of urea = 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 gMass of urea required in making 25 grams of 0.25 m = 37 grams

Explanation:

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