Expert Chemistry Tutoring

MS1:

- Concentration is the amount of solute in a given volume of solution. (1)

- Units: mol/dm³ or g/dm³ (1)

MS2:

- Use the dilution formula: C₁V₁ = C₂V₂ (1)

- Calculate the volume of stock solution needed. (1)

- Add the calculated volume of stock solution to water and mix. (1)

MS3:

- Moles of NaCl = 0.1 × 0.25 = 0.025 mol (1)

- Mass of NaCl = moles × relative formula mass = 0.025 × 58.5 = 1.4625 g (3)

MS4:

- Molarity (M) is moles of solute per cubic decimeter of solution. (1)

- Molality (m) is moles of solute per kilogram of solvent. (1)

MS5:

- Concentration of reactants (1)

- Surface area of reactants (1)

- Temperature (1)

MS6:

- Moles of CuSO₄ = mass ÷ relative formula mass = 2.5 ÷ 160 = 0.0156 mol (1)

- Concentration = moles ÷ volume = 0.0156 ÷ 0.1 = 0.156 mol/dm³ (3)

MS7:

- A limiting reactant is the reactant that is used up first in a chemical reaction. (2)

MS8:

- From the balanced equation, 2 moles of Na produce 2 moles of NaCl. (1)

- Mass of NaCl = moles × relative formula mass = 0.5 × 58.5 = 29.25 g (2)

MS9:

- Moles of HCl = concentration × volume = 1 × 0.03 = 0.03 mol (1)

- From the balanced equation, 1 mole of NaOH neutralizes 1 mole of HCl. (1)

- Moles of NaOH = 0.03 mol (2)

- Volume of NaOH solution = moles ÷ concentration = 0.03 ÷ 2 = 0.015 dm³ or 15 cm³ (3)

MS10:

- % mass = (mass of solute ÷ mass of solution) × 100 (1)

- Mass of solution = mass of solute + mass of solvent (1)

- Example: A solution contains 5 g of salt dissolved in 50 g of water.

- % mass of salt = (5 ÷ 55) × 100 = 9.09% (1)