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

116 mL of 0.480 M KCl

4.15

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Correct

Part B

150 g of 0.600 m KCl

6.70

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Part C

150 g of 5.0 % KCl solution by mass

Express your answer using two significant figures.

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Part D

How much solvent would you take to prepare the solution in part B?

Express your answer using four significant figures.

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Part E

How much solvent would you take to prepare the solution in part C?

Express your answer using three significant figures.

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

Part A

116 mL of 0.480 M KCl

msolute =

4.15

gKCl

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Correct

Part B

150 g of 0.600 m KCl

msolute =

6.70

gKCl

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Incorrect; Try Again; 5 attempts remaining

Part C

150 g of 5.0 % KCl solution by mass

Express your answer using two significant figures.

msolute = gKCl

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Part D

How much solvent would you take to prepare the solution in part B?

Express your answer using four significant figures.

msolvent = gH2O

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Part E

How much solvent would you take to prepare the solution in part C?

Express your answer using three significant figures.

msolvent = gH2O

Explanation / Answer

molarity   = mass of solute * 1000/gram molar mass of solute * volume of solution in ml

A. 0.48   = mass of KCl*1000/74.5*116

mass of KCl = 0.48*74.5*116/1000 = 4.15g

B.

molality   = mass of solute *1000/gram molar mass of solute * mass of solvent

0.6        = mass of solute *1000/74.5*150

mass of solute Kcl = 0.6*74.5*150/1000 = 6.705g

C.

5% KCl solution means

5g of KCl present in 100g of solution

100g of solution contains 5g of KCl

150g of solution contains = 5*150/100 = 7.5g of KCl

D. 150g of H2O

E . solution = solute + solvent

     150g     = 7.5g + solvent

mass of solvent = 150-7.5 = 142.5g of H2O

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