Question #b8448

1 Answer
Dec 9, 2015

Answer:

#"14.7 mL"#

Explanation:

Your strategy here will be to

  • use the molarity and volume of the target solution to find how many moles of solute it must contain

  • use cesium chloride's molar mass to determine how many grams of cesium would contain that many moles

  • use the stock solution's known percent concentration by mass to determine what mass of the stock solution would contain that many grams of cesium

  • use the stock solution's density to determine the volume of solution that would contain that many grams of solution

As you know, molarity is defined as moles of solute, which in your case is cesium chloride, #"CsCl"#, divided by Liters* of solution. This means that the number of moles of cesium chloride that must be present in the target solution is equal to

#color(blue)(c = n/V implies n = c * V)#

#n = "0.100 M" * 500.0 * 10^(-3)"L" = "0.0500 moles CsCl"#

Use the compound's molar mass to help you find the mass of cesium chloride that would contain this many moles

#0.0500 color(red)(cancel(color(black)("moles CsCl"))) * "168.37 g CsCl"/(1color(red)(cancel(color(black)("mole CsCl")))) = "8.4185 g CsCl"#

Now, the stock solution is #"40.0% w/w"#, which means that every #"100 g"# of solution would contain #"40.0 g"# of cesium chloride.

This means that the mass of stock solution that would contain this many grams of cesium chloride is equal to

#8.4185 color(red)(cancel(color(black)("g CsCl"))) * "100.0 g stock"/(40.0color(red)(cancel(color(black)("g CsCl")))) = "21.046 g stock"#

You know that this stock solution has a density of #"1.43 g/mL"#, which means that every milliliter of solution has a mass of #"1.43 g"#.

In your case, the volume that contains #21.046# grams of stock solution will be

#21.046 color(red)(cancel(color(black)("g stock"))) * "1 mL"/(1.43color(red)(cancel(color(black)("g stock")))) = "14.717 mL"#

Rounded to three sig figs, the answer will be

#V_"stock" = color(green)("14.7 mL")#

So, to prepare your target solution, take #"14.7 mL"# of the stock solution and add enough water to make the volume of the solution equal to #"500.0 mL"#.