answer this question please
Answers
Answer:
66.67% is the mass percent of water
1.46 m is the molality of the solution.
Explanation:
Refer attachment for explanation!!!
Answer- The above question is from the chapter 'Some Basic Concepts of Chemistry'.
·Mass % of a solute/solvent= Mass of solute/solvent÷Mass of compound × 100
·Molality- It is the number of moles of solution in 1 kg of solvent.
Molality = Moles of solute ÷ Mass of solvent in kgs
Given question: A solution is prepared by mixing 50 g sugar in 100 g of water at 25 °C. Calculate the following:
a) Mass percent of water.
b) Molality of the solution.
Answer: Mass of solute (sugar) = 50 g
Mass of solvent (water) = 100 g
Mass of solution = Mass of solute + Mass of solvent = 50 + 100 = 150 g
a) Mass percent of water = Mass of water/Mass of solution × 100
= 100/150 × 100 = 66.67%
b) Moles of solute = Given mass/Molar mass
= 50/342 moles
Mass of solvent in kg = 1000/100 kg
Molality = Moles of solute ÷ Mass of solvent in kgs
= 50/342 × 1000/100
= 1.461 molal = 1.416 m
∴ Mass % of water = 66.67 %
Molality of solution = 1.416 m