If you dissolve 0.0147 kg of glucose (c6h12o6) using 0.150 l of water with a density of 1.00 g/ml, what will be the molality?

Question

Hillary Roberts

Chemistry

7 December, 02:26

If you dissolve 0.0147 kg of glucose (c6h12o6) using 0.150 l of water with a density of 1.00 g/ml, what will be the molality?

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Answers (1)

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Clare Mahoney · Answer 1

Moles of glucose = 14.7/180.15 = 0.0816moles.

Mass of solvent (water) = density of water x volume = 1 x 150 g = 150g

Mass of water in Kg = 0.15 Kg

Molality of glucose = moles of glucose / mass of water

= 0.0816 / 0.15 = 0.54 molality.