23.12 grams of glucose (C6H12O6) is dissolved in 1.00 kg of water. Calculate the molality
1. we find the amount of matter (mol) of glucose. To do this, we divide the given mass of glucose (180g/mol) by its molecular mass:
n = m / M = 23.12 / 180 = 0.1284 mol.
2. we find the molal concentration. To do this, we divide the amount of substance (n, mol) of glucose by the mass of the solvent, is water (kg):
Cmolality = 0.1284 mol / 1 kg = 0.1284
ANSWER: molality=0.1284
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