Solved 3. A0.750 M solution of HSO, in water has a density | Chegg.com - Free Printable

You're working on a multi-part concentration problem for a 0.750 M H₂SO₄ solution with density = 1.046 g/mL at 20°C, and molar mass of H₂SO₄ = 98.1 g/mol.

Let’s solve each part step-by-step as requested:

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## ✔ ASSUMPTION: We assume 1.00 L of solution (since molarity is per liter).

This is standard practice for these types of problems — it simplifies calculations.

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## 🔹 (a) Mole Fraction of H₂SO₄

Mole fraction (χ) = moles of solute / total moles in solution

We need:
- Moles of H₂SO₄
- Moles of H₂O (solvent)

Step 1: Moles of H₂SO₄

Given: 0.750 M → so in 1 L, we have 0.750 mol H₂SO₄

✔ Moles of H₂SO₄ = 0.750 mol

Step 2: Mass of Solution

Density = 1.046 g/mL → for 1 L = 1000 mL

→ Mass of solution = 1000 mL × 1.046 g/mL = 1046 g

Step 3: Mass of H₂SO₄

Mass = moles × molar mass = 0.750 mol × 98.1 g/mol = 73.575 g

Step 4: Mass of Water (Solvent)

Mass of water = mass of solution – mass of solute
= 1046 g – 73.575 g = 972.425 g

Step 5: Moles of Water

Molar mass H₂O = 18.0 g/mol
→ Moles of H₂O = 972.425 g ÷ 18.0 g/mol ≈ 54.0236 mol

Step 6: Total Moles in Solution

Total moles = moles H₂SO₄ + moles H₂O = 0.750 + 54.0236 ≈ 54.7736 mol

Step 7: Mole Fraction of H₂SO₄

χ(H₂SO₄) = 0.750 / 54.7736 ≈ 0.0137

✔ Answer (a): Mole Fraction = 0.0137

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## 🔹 (b) Mass Percent of H₂SO₄

Mass percent = (mass of solute / mass of solution) × 100%

= (73.575 g / 1046 g) × 100% ≈ 7.03%

✔ Answer (b): Mass Percent = 7.03%

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## 🔹 (c) Molality of H₂SO₄

Molality (m) = moles of solute / kg of solvent

We already have:
- Moles of H₂SO₄ = 0.750 mol
- Mass of solvent (water) = 972.425 g = 0.972425 kg

→ Molality = 0.750 mol / 0.972425 kg ≈ 0.771 m

✔ Answer (c): Molality = 0.771 m

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## 🧾 FINAL ANSWERS:

> (a) Mole Fraction = 0.0137
> (b) Mass Percent = 7.03%
> (c) Molality = 0.771 m

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## 💡 Notes:

- You correctly assumed 1 L of solution — great start!
- You were on the right track for mass percent and molality.
- For mole fraction, you needed to find moles of water — which requires converting mass of water to moles using 18.0 g/mol.
- All steps are consistent and use given values accurately.

Let me know if you’d like to see how this changes if you used a different volume or want to verify with another method!