Question

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

Part B

112g of 0.425m KCl

Express your answer using three significant figures.msolute=gKCL

Part D

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

Express your answer using four significant figures. msolvent=gH20

Part E

How much solvent would you take to prepare the solution in part C? Part C:112g of 4.2% KCl solution by mass ANSWER:msolute= 4.7gKCL

Express your answer using two significant figures. msolvent gH20

Answer 1

Part B: Mass of KCl

Let x be moles of KCl and y the mass of solvent (in kg):

The mass of the solution is the sum of the mass of solute and the mass of solvent, then:

The molecular weight of KCl is 74.544 g/mol, therefore:

Substituting the first equation into the equation above:

And, solving the above equation for x:

So, the mass of solute is:

Part D: Mass of H₂O

The mass of H₂O can be calculated bubstituting x=0.0461 into the following equation, and solving for y, :

Part E: Mass of H₂O

The mass of the solution is the sum of the mass of solute and the mass of solvent, then: