Question

23.12 grams of glucose (C6H12O6) is dissolved in 1.00 kg of water. Calculate the molality

Answer 1

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):

C_molality = 0.1284 mol / 1 kg = 0.1284

ANSWER: molality=0.1284