Calculate the molality of C2H5OH in a water solution that is prepared by mixing 50.0 mL of C2H5OH with 112.7 mL of H2O at 20°C. The density of the C2H5OH is 0.789 g/mL at 20°C. (Assume the density of water at this temperature is 1.00 g/mL.)

Question

Chemistry

Arroyo

13 June, 13:28

Calculate the molality of C2H5OH in a water solution that is prepared by mixing 50.0 mL of C2H5OH with 112.7 mL of H2O at 20°C. The density of the C2H5OH is 0.789 g/mL at 20°C. (Assume the density of water at this temperature is 1.00 g/mL.)

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Adelyn · Answer 1

C C2H5OH = 7.598 molal

Explanation:

molality (m) C2H5OH = n C2H5OH / Kg H2O

∴ V C2H5OH = 50.0 mL

∴ V H2O = 112.7 mL

∴ T = 20 °C

∴ δ C2H5OH (20°C) = 0.789 g/mL

∴ δ H2O (20°C) = 1.00 g/mL

∴ Mw C2H5OH = 46.0684 g/mol

⇒ n C2H5OH = (50.0 mL) * (0.789 g/mL) * (mol/46.0684 g) = 0.8563 mol

⇒ mass H2O = (112.7 mL) * (1.00 g/mL) * (Kg / 1000 g) = 0.1127 Kg

⇒ C C2H5OH = 0.8563 mol C2H5OH / 0.1127 Kg H2O

⇒ C C2H5OH = 7.598 molal