calculate W/W, V/V, W/V, mole fraction of solute and solvent and also calculate molarity, molality.weight of C6H6 (benzene)19.5gm, weight of C6H6O (acetone) 15gm.(density of bezene=1.2g/mol, density of acetone 1.5g/mol)

Answer 1

To calculate the quantities requested, we need to know the total volume of the solution. Let's assume that we have mixed 50mL of benzene (density=1.2 g/mL) with 50mL of acetone (density=0.789 g/mL).

To calculate W/W:
- Weight of solute (C6H6O) = 15g
- Weight of solvent (C6H6) = 19.5g
- Total weight of solution = 15g + 19.5g = 34.5g
- W/W of solute = (15g / 34.5g) x 100% = 43.48%
- W/W of solvent = (19.5g / 34.5g) x 100% = 56.52%

To calculate V/V:
- Volume of solute (acetone) = 50mL
- Volume of solvent (benzene) = 50mL
- Total volume of solution = 50mL + 50mL = 100mL
- V/V of solute = (50mL / 100mL) x 100% = 50%
- V/V of solvent = (50mL / 100mL) x 100% = 50%

To calculate W/V:
- Weight of solute (C6H6O) = 15g
- Volume of solvent (benzene) = 50mL
- W/V of solute = (15g / 50mL) x 100% = 30%
- Weight of solvent (C6H6) = 19.5g
- Volume of solvent (benzene) = 50mL
- W/V of solvent = (19.5g / 50mL) x 100% = 39%

To calculate mole fraction:
- Moles of solute (C6H6O) = 15g / 58.08 g/mol = 0.258 moles
- Moles of solvent (C6H6) = 19.5g / 78.11 g/mol = 0.25 moles
- Total moles of solution = 0.258 moles + 0.25 moles = 0.508 moles
- Mole fraction of solute = 0.258 moles / 0.508 moles = 0.508
- Mole fraction of solvent = 0.25 moles / 0.508 moles = 0.492

To calculate molarity:
- Volume of solution = 100mL = 0.1L
- Moles of solute (C6H6O) = 0.258 moles
- Molarity of solute = 0.258 moles / 0.1 L = 2.58 M

To calculate molality:
- Weight of solvent (C6H6) = 19.5g
- Volume of solvent (benzene) = 50mL = 0.05 L
- Moles of solvent (C6H6) = 19.5g / 78.11 g/mol = 0.25 moles
- Molality of solution = 0.258 moles / 0.05 kg = 5.16 mol/kg

To calculate weight of C6H6 (benzene) and C6H6O (acetone):
- Weight of C6H6 (benzene) = 50mL x 1.2 g/mL = 60 g
- Weight of C6H6O (acetone) = 15g