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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 !!!!,

User Nilloc
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1 Answer

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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
User Hrafn
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