Respuesta :

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