Question

Question #b54e2

Answer 1

`"198 g"`

Explanation:

Your strategy here will be to use the target solution's molality to determine how many moles of strontium chloride it must contain, then use the compound's molar mass to convert the number of moles to grams.

Molality is a measure of a solution's concentration in terms of moles of solute and kilograms of solvent. More specifically, a solution's molality tells you how many moles of solute you get per kilogram of solvent.

This means that a `"0.500 m"` solution will contain `0.500` moles of strontium chloride, your solute, per kilogram of water, your solvent.

You can thus use the molality of the target solution as a conversion factor to help you calculate how many moles of solute would be present in `2.50` kilograms of solvent

`2.50 color(red)(cancel(color(black)("kg water"))) * overbrace("0.500 moles SrCl"_2/(1color(red)(cancel(color(black)("kg water")))))^(color(purple)("= 0.500 m")) = "1.250 moles SrCl"_2`

Now all you have to do is use strontium chloride's molar mass as a conversion factor to determine how many grams would contain that many moles

`1.250 color(red)(cancel(color(black)("moles SrCl"_2))) * "158.53 g"/(1color(red)(cancel(color(black)("mole SrCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)"198 g"color(white)(a/a)|)))`

The answer is rounded to three sig figs.