Question #31da1

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Question #31da1

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Answer 1 · 2016-03-18T19:59:55

27.745 g of calcium chloride

Explanation:

Molarity is number of moles of solute per litre of solution. You need to determine the number of grams of solute, so first you need to know the number of moles.

Rearranging Molarity = moles / litres gives Moles = Litres x molarity.

125 g of water is equivalent to 125 ml, which is 0.125 litre. Molarity is 2.0M, so the number of moles is 0.125 x 2 = 0.25 mol.

Now simply multiply this by the molar mass of calcium chloride, which is 110.98 g/mol, giving you the answer 27.745 g.

Answer 2 · 2016-03-19T20:14:02

${28 g CaCl}_{2}$ $"28 g CaCl"_2$

Explanation:

A solution's molality will tell you how many moles of solute, which in your case is calcium chloride, ${CaCl}_{2}$ $"CaCl"_2$ , you get per kilogram of solvent.

Basically, molality uses moles of solute and mass of solvent to express the concentration of a given solution.

$∣ ∣ ∣ ∣ ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ \frac{a}{a} molality = \frac{moles of solute}{kilogram of solution} \frac{a}{a} ∣ ∣ ∣ - ------------------------------ -$ $color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solution"color(white)(a/a)|)))$

So, you know that your target solution must have a molaity of $2.0 m$ $"2.0 m"$ , or ${2.0 mol kg}^{- 1}$ $"2.0 mol kg"^(-1)$ . This means that every kilogram of water, which is the solvent, will contain $2.0$ $2.0$ moles of calcium chloride.

Now, you know that you only have $125 g$ $"125 g"$ of water to make this solution. You can use the molality of the target solution as a conversion factor to help you determine how many moles of calcium chloride must be dissolved in $125 g$ $"125 g"$ of water in order to have a ${2.0 mol kg}^{- 1}$ $"2.0 mol kg"^(-1)$ solution.

Keep in mind that molality uses kilograms of solvent, so use the conversion factor

$1 kg = 10^{3} g$ $"1 kg" =10^3"g"$

to go from grams of water to kilograms of water. You will thus have

$125 \cdot 10^{- 3} kg water \cdot equivalent to 2.0 m      \frac{{2.0 moles CaCl}_{2}}{1 kg water} = {0.250 moles CaCl}_{2}$ $125 * 10^(-3)color(red)(cancel(color(black)("kg water"))) * overbrace("2.0 moles CaCl"_2/(1color(red)(cancel(color(black)("kg water")))))^(color(purple)("equivalent to 2.0 m")) = "0.250 moles CaCl"_2$

To find how many grams of calcium chloride would contain this many moles, use the compound's molar mass

$0.250 {moles CaCl}_{2} \cdot \frac{111 g}{1 {mole CaCl}_{2}} = 27.75 g$ $0.250color(red)(cancel(color(black)("moles CaCl"_2))) * "111 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = "27.75 g"$

Rounded to two sig figs, the number of sig figs you have for the molality of the target solution, the answer will be

${mass of CaCl}_{2} = ∣ ∣ ∣ ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ \frac{a}{a} 28 g \frac{a}{a} ∣ ∣ - ------ -$ $"mass of CaCl"_2 = color(green)(|bar(ul(color(white)(a/a)"28 g"color(white)(a/a)|)))$

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Question #31da1

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