Sample Exercise 4: Calculation of molality and Molarity A solution is made by dissolving 4.35 g glucose (C6H₁2O6) in 25.0 mL of water at 25 °C. Calculate the (a) molality and (b) the Molarity of glucose in the solution. Water has a density of 1.00 g/mL. The glucose solution has a density of 1.046 g/mL. Solution (a) Calculating molality requires that we know the mass of both components separately. We then use the MM of the solute to get the moles of solute. mass H₂O: = 25.0 g H₂O MM C6H₁2O6 = 180.16 g/mol 25.0 mL x 1.00 g =

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Sample Exercise 4: Calculation of molality and Molarity
A solution is made by dissolving 4.35 g glucose (C6H₁2O6) in 25.0 mL of water at 25 °C.
Calculate the (a) molality and (b) the Molarity of glucose in the solution. Water has a
density of 1.00 g/mL. The glucose solution has a density of 1.046 g/mL.
Solution
(a) Calculating molality requires that we know the mass of both components
separately. We then use the MM of the solute to get the moles of solute.
mass H₂O: = 25.0 mL x 1.00 g
25.0 mL x
mL
moles glucose:
1.00 g = 25.0 g H₂O MM C6H₁₂O6 = 180.16 g/mol
4.35 g
180.16 g mol-¹
= 0.024145 mol C6H₁2 06
molality glucose: 0.024145 mol = 0.966 m
0.0250 kg H₂O
Molarity glucose:
(b) To get the Molarity of the glucose in the solution, we need to know the mass of
the solution, then use the solution density to find the volume of the solution.
mass of solution: mass H₂O + mass glucose = 25.0 g + 4.35 g = 29.35 g
volume of solution: 29.35 g x 1 mL = 28.06 mL of sol'n
1.046 g
= 0.860 M
0.024145 mol
0.02806 L sol'n
Transcribed Image Text:Sample Exercise 4: Calculation of molality and Molarity A solution is made by dissolving 4.35 g glucose (C6H₁2O6) in 25.0 mL of water at 25 °C. Calculate the (a) molality and (b) the Molarity of glucose in the solution. Water has a density of 1.00 g/mL. The glucose solution has a density of 1.046 g/mL. Solution (a) Calculating molality requires that we know the mass of both components separately. We then use the MM of the solute to get the moles of solute. mass H₂O: = 25.0 mL x 1.00 g 25.0 mL x mL moles glucose: 1.00 g = 25.0 g H₂O MM C6H₁₂O6 = 180.16 g/mol 4.35 g 180.16 g mol-¹ = 0.024145 mol C6H₁2 06 molality glucose: 0.024145 mol = 0.966 m 0.0250 kg H₂O Molarity glucose: (b) To get the Molarity of the glucose in the solution, we need to know the mass of the solution, then use the solution density to find the volume of the solution. mass of solution: mass H₂O + mass glucose = 25.0 g + 4.35 g = 29.35 g volume of solution: 29.35 g x 1 mL = 28.06 mL of sol'n 1.046 g = 0.860 M 0.024145 mol 0.02806 L sol'n
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