Answer:
Molarity → 0.410 M
Molality → 0.44 m
Percent by mass → 7.61 g
Mole fraction (Xm) = 7.96×10⁻³
Mole percent = 0.79 %
Explanation:
We analyse data:
28.8 g of glucose → To determine molarity, molality, mole fraction, mole percent we need to find out the moles:
28.2 g. 1mol / 180 g = 0.156 moles of glucose
350 g of water → Mass of solvent.
We convert from g to kg in order to determine molality = 350 g / 1000 = 0.350 kg
We also need the moles of solvent: 350 g / 18 g/mol = 19.44 moles
380 mL of solution → Volume of solution; to determine the molarity we need the volume in L → 380 mL / 1000 = 0.380L
Solution mass = Solute mass + Solvent mass
28.2 g + 350 g = 378.2 g
Total moles = Moles of solute + Moles of solvent
0.156 + 19.44 = 19.596 moles
Molarity → Moles of solute in 1L of solution → 0.156 mol / 0.380L = 0.410 M
Molality → Moles of solute in 1kg of solvent → 0.156 mol / 0.350 kg = 0.44 m
Percent by mass → Mass of solute in 100 g of solution
(28.8g /378.2g) . 100 = 7.61 g
Mole fraction (Xm)= Moles of solute/ Total moles → 0.156 mol / 19.596 moles = 7.96×10⁻³
Mole percent = Xm . 100 → 7.96×10⁻³ . 100 = 0.79 %
The molarity of the solution is 0.42 M. The molality of the solution is 0.46 m.
a)To obtain the molarity of the solution;
Number of moles = mass of glucose/molar mass of glucose = 28.8g/180 g/mol
= 0.16 moles
Volume of solution = 380mL or 0.38 L
Molarity = number of moles /volume = 0.16 moles/ 0.38 L = 0.42 M
b) Molality of the solution;
Mass of solvent in Kg = 350g/1000 = 0.35 Kg
Molality = number of moles/mass of solution in kilogram = 0.16 moles/ 0.35 Kg = 0.46 m
c) percent by mass = mass of solute/mass of solution× 100
= 28.8g/(350g +28.8g) × 100 = 7.6 %
d) Mole fraction
Number of moles of water = 350g/ 18 g/mol = 19.44 moles
Total number of moles = 19.44 moles + 0.16 moles = 19.6 moles
Mole fraction of glucose = 0.16 moles/19.6 moles = 0.0082
e) Mole percent
Mole fraction × 100 = 0.0082 × 100 = 0.82%
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