Molarity and Molality

Chemistry Level pending

If 8 8 g of sodium hydroxide (NaOH) is dissolved in 100 100 g of water, the density of the solution is 1.2 1.2 g/mL. What are the approximate molality and molarity (molar concentration) of this solution? (The formula weight of NaOH is 40 40 .)

2.0 mol/kg - 2.0 mol/L 2.0 mol/kg - 2.2 mol/L 2.2 mol/kg - 2.2 mol/L 2.2 mol/kg - 2.0 mol/L

Lakshan Dumpa by Lakshan Dumpa , 24, India

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1 solution

Lu Chee Ket
Feb 1, 2016

Molality = 8 40 m o l 100 g = 2 m o l 1 k g = 2.0 m o l / k g \Large \frac{\frac{8}{40} mol}{100~g} = \frac{2~mol}{1~kg} = 2.0~mol/ kg = 2.0 m

0.2 mol in 108 g solution = 0.2 mol in 108 g 1.2 g / m l \frac{108~g}{1.2~g/ ml} = 0.2 mol in 90 ml

0.2 m o l 90 m l = 0.2 m o l × 1000 m l 90 m l × 1000 m l = 2.222 m o l 1 L \frac{0.2~mol}{90~ml} = \frac{0.2~mol \times 1000~ml}{90~ml \times 1000~ml} = \frac{2.222~mol}{1~L}

Molarity ≈ 2.2 m o l / L \approx 2.2~mol/ L = 2.2 M

Answer: 2.0 m o l / k g − 2.2 m o l / L \boxed{2.0~mol/ kg - 2.2~mol/ L}

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