A solution contains 48.6 g glucose (c6h12o6) dissolved in 0.800 l of water. what is the molality of the solution? (assume a density of 1.00 g/ml for water.)

Question

Gracie Sexton

Chemistry

7 October, 01:41

A solution contains 48.6 g glucose (c6h12o6) dissolved in 0.800 l of water. what is the molality of the solution? (assume a density of 1.00 g/ml for water.)

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Ricardo Meadows · Answer 1

Answer is: molality is 0,3375 mol/kg.

m (C₆H₁₂O₆) = 48,6 g.

V (H₂O) = 0,800 L · 1000 mL/1L = 800 mL.

d (H₂O) = 1 g/mL.

m (H₂O) = V (H₂O) · d (H₂O).

m (H₂O) = 800 mL · 1 g/mL.

m (H₂O) = 800 g : 1000 g/1kg = 0,8 kg.

n (C₆H₁₂O₆) = m (C₆H₁₂O₆) : M (C₆H₁₂O₆).

n (C₆H₁₂O₆) = 48,6 g : 180,15 g/mol.

n (C₆H₁₂O₆) = 0,27 mol.

b (C₆H₁₂O₆) = n (C₆H₁₂O₆) : m (H₂O).

b (C₆H₁₂O₆) = 0,27 mol : 0,8 kg = 0,3375 mol/kg.