How much dry solute would you take to prepare each of the following solutions fr

Question

How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent?

part a.) 133mL of 0.535M KCl Express your answer using three significant figures.

part b.) 127g of 0.450m KCl Express your answer using three significant figures.

part c.) 127g of 5.4% KCl solution by mass. Express your answer using two significant figures.

part d.) How much solvent would you take to prepare the solution in part B? Express your answer using four significant figures.

part e.) How much solvent would you take to prepare the solution in part C? Express your answer using two significant figures.

Explanation / Answer

a)

Moles of KCl = 133 mL x 1 L/1,000 mL x 0.535 moles/L = 0.0712 moles

Mass of KCl = 0.0712 moles x 74.55 g/mol = 5.31 g

b)

0.450 m = 0.450 moles of solute / kg of solvent

Mass of KCl = 0.450 moles x 74.55 g/mol = 33.5 g

Mass of solute + mass of solvent = mass of solution

Mass of solution = 33.5 g + 1,000 g = 1,033.5 g

Mass of KCl = 127 g of solution x (33.5 g of KCl / 1,033.5 g of solution) = 4.12 g

c)

5.4% KCl by mass = 5.4 g of KCl / 100 g of solution

Mass of KCl = 127 g of solution x (5.4 g of KCl / 100 g of solution) = 6.9 g

d)

Mass of solute + mass of solvent = mass of solution

Mass of solvent = mass of solution - mass of solute = 127 g - 4.12 g = 122.9 g

e)

Mass of solute + mass of solvent = mass of solution

Mass of solvent = mass of solution - mass of solute = 127 g - 6.9 g = 1.2 x 10^2 g

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