Question

what is the molality of a 10.5% by mass glucose (C6H12O6) the density of the solution is 1.03 g/ml. i got up to m = 0.058281527 mol/ kg... please help !!!!,

Answer 1

Solution:
1 L of solution and 1.03 g/mL = 1030 g/L 

1030 g/L time 10.5%
= 108.15 g of glucose/L
And
=108.15 g / 180.16 g/mol
= 0.6003 moles in 1L 
=or 0.600 mol/kg