Answer:
[Na₂S₂O₃] = 0.83 M
[Na₂S₂O₃] = 0.86 m
Mole fraction = 0.015
Explanation:
Na₂S₂O₃ 12 % by mass. This data means, that 12 g of solute are contained in 100 g of solution.
Let's find out the volume of solution, with density to determine molarity.
Solution density = Solution mass / Solution volume
1.1003 g/cm³ = 100 g / Solution volume
100 g / 1.1003 g/cm³ = Solution volume → 90.88 mL (1cm³ = 1mL)
Now, that we have volume, we can calculate molarity
Molarity is mol/L
90.88 mL = 0.09088 L
12 g / 158.12 g/mol = 0.0759 moles
0.0759 moles / 0.09088 L = 0.83 M
Total mass of solution = 100 g
12 g + Solvent mass = 100 g
Solvent mass = 100 g - 12 g → 88 g
Molality = moles of solute /1kgof solvent
88 g = 0.088 kg
0.0759 moles / 0.088 kg = 0.86 m
As solvent mass is 88 g, let's determine solvent's moles for mole fraction
88 g / 18 g/mol = 4.89 moles
Mole fraction = moles of solute / moles of solutes + moles of solvent
Mole fraction = 0.0759 mol / 0.0759 mol + 4.89 moles = 0.015
