Question

Question #58a1e

Answer 1

You'd need to add 6.11 g of potassium chloride to get that respective molarity.

Molarity is defined as moles of solute, in your case potassium chloride, per liters of solution. Potassium chloride is a very soluble salt, so you can assume that the total volume of the solution will be equal to the volume of the water.

You can use the density and mass of the water to determine the volume the solution would have

`rho = m/V => V = m/(rho)`

`V_"sol" = (372cancel("g"))/(0.998cancel("g")/"cm"^3) = "372.75 mL"`

SInce you know what the molarity of the target solution must be, you can use the volume of the solution to determine how many moles of solute would be needed to get that respective molarity

`C = n/V => n = C * V`

`n_(KCl) = "0.220 M" * 372.75 * 10^(-3)"L" = "0.0820 moles"` `KCl`

Now use potassium chloride's molar mass to determine how many grams would contain this many moles

`0.0820cancel("moles") * "74.55 g"/(1cancel("mole")) = "6.113 g"`

Rounded to three sig figs, the answer will be

`m_(KCl) = color(green)("6.11 g")`