What is the molality of a solution prepared by dissolving 225 mg of glucose (C6H12O6) in 5.00 mL ofethanol (density = 0.789 g/mL)
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What is the molality of a solution prepared by dissolving 225 mg of glucose (C6H12O6) in 5.00 mL ofethanol (density = 0.789 g/mL)
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Concept: Molality is the amount of a substance dissolved in a certain mass of solvent.
Find: Calculate the molality of a solution.
Solution:
Molality=[tex]\frac{amount of solute}{mass of solvent}[/tex]
Determine moles of [tex]C_{6} H_{12} 0_{6}[/tex]
mol solute=150.0g X[tex]\frac{1 molC_{6} H_{12} 0_{6} }{180.15588gC_{6} H_{12} 0_{6}}[/tex]
= 0.8326 mol [tex]C_{6} H_{12} 0_{6}[/tex]
Convert mass of solvent from g to kg.
=600.0g [tex]H_{2} o[/tex]X[tex]\frac{1 Kg H_{2} o}{1000gH_{2} o}[/tex]
=0.6000 kg [tex]H_{2} o[/tex]
Calculate molality:
molality=[tex]\frac{0.8326mol}{0.6000kg}[/tex]
=1.388 mol/kg
= 1.388 m
Final answer: Molality of a solution is 1.388m.
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