# Question d232c

Jan 16, 2017

#### Answer:

Here's what I got.

#### Explanation:

I'll show you how to solve the first problem.

Let's start with part (a).

So, you know that you have a ${\text{0.4 mol dm}}^{- 3}$ solution of fructose that you must prepare in a total volume of ${\text{35 cm}}^{- 3}$ of solution.

The first thing to do here is to figure out how many moles of fructose must be dissolved in water in order to get ${\text{35 cm}}^{- 3}$ of ${\text{0.4 mol dm}}^{- 3}$ solution.

Use the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{1 dm"^3 = 10^3"cm}}^{3}}}}$

to calculate the number of moles of fructose

35 color(red)(cancel(color(black)("cm"^3 "solution"))) * (1color(red)(cancel(color(black)("dm"^3))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * "0.4 moles fructose"/(1color(red)(cancel(color(black)("dm"^3 "solution"))))

$= \text{ 0.0140 moles fructose}$

To convert this to grams of fructose, use the molar mass of the compound

$0.0140 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles fructose"))) * "180 g"/(1color(red)(cancel(color(black)("mole fructose")))) = color(darkgreen)(ul(color(black)("2.5 g}}}} \to$ two sig figs

For part (b), you know that you have a ${\text{2.4 mol dm}}^{- 3}$ stock solution that must be used to prepare ${\text{375 cm}}^{3}$ of ${\text{0.4 mol dm}}^{- 3}$ solution.

Notice that the concentration of the diluted solution is $6$ times lower than the concentration of the stock solution

"DF" = (2.4 color(red)(cancel(color(black)("mol dm"^(-3)))))/(0.4color(red)(cancel(color(black)("mol dm"^(-3))))) = color(blue)(6) -> the dilution factor

This means that the volume of the diluted solution must be $6$ timer larger than the volume of the stock solution.

$\text{DF" = V_"diluted"/V_"stock" implies V_"stock" = V_"diluted"/"DF}$

Keep in mind that the dilution factor, $\text{DF}$, essentially tells you how concentrated the stock solution was compared with the diluted solution.

You will thus have

${V}_{\text{stock" = "375 cm"^3/color(blue)(6) = "63 cm}}^{3} \to$ two sig figs

So, in order to prepare this solution, you would start with ${\text{63 cm}}^{3}$ of ${\text{2.4 mol dm}}^{- 3}$ solution and add enough water to get the total volume of the solution to ${\text{375 cm}}^{3}$.

For part (c)), I'll assume that you must express the concentration of the ${\text{0.4 mol dm}}^{- 3}$ fructose solution in milligrams per $100$ cubic centimeters, ${\text{mg cm}}^{- 3}$.

You already know from part (a) that

"0.4 mol fructose"/("1 dm"^3 "solution") = "2.5 g fructose"/("35 cm"^3"solution")

so start by calculating how many grams of fructose you have per ${\text{100 cm}}^{3}$ of solution

100 color(red)(cancel(color(black)("cm"^3"solution"))) * "2.52 g fructose"/(35color(red)(cancel(color(black)("cm"^3"solution")))) = "7.14 g fructose"

This means that you have

"0.4 mol fructose"/("1 dm"^3 "solution") = "7.14 g fructose"/("100 cm"^3"solution")

Finally, use the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 g" = 10^3"mg}}}}$

to get

(7.14 color(red)(cancel(color(black)("g"))) "fructose")/"100 cm"^3 * (10^3"mg")/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("71 mg / 100 cm"^3))) -> two sig figs

I'll leave the values rounded to two sig figs, but keep in mind that you only have one sig figs for the initial concentration of the solution.

Jan 17, 2017

#### Answer:

Here's what I got for the answers to Part b) of your second question:

0.011 mol/L; 11 mmol/L; 83 mg%; 0.56 ‰.

#### Explanation:

You are starting with 10 g of $\text{KCl}$ in $\text{12 L}$ of a solution with a density of $\text{1.5 g/mL}$.

1. Molarity

$\text{Molarity" = "moles"/"litres}$

$\text{Moles of KCl" = 10 color(red)(cancel(color(black)("g KCl"))) × "1 mol KCl"/(74.5 color(red)(cancel(color(black)("mol KCl")))) = "0.134 mol KCl}$

$\text{Molarity" = "0.134 mol"/"12 L" = "0.011 mol/L}$

2. Millimolarity

Here, we must use the conversion $\text{1 mol = 1000 mmol}$.

$\text{Millimolarity" = (0.011 color(red)(cancel(color(black)("mol"))))/("1 L") × "1000 mmol"/(1 color(red)(cancel(color(black)("mol")))) = "11 mmol/L}$

3. Mg percentage

The term "mg%" means the mass in milligrams of a substance in 100 mL of solution.

You have 10 g of $\text{KCl}$ in 12 L of solution.

∴ In 100 mL of solution, you have

$\text{ Mass of KCl" = 100 color(red)(cancel(color(black)("mL solution"))) × (10 color(red)(cancel(color(black)("g KCl"))))/("12 000" color(red)(cancel(color(black)("mL solution")))) × "1000 mg KCl"/(1 color(red)(cancel(color(black)("g KCl")))) = "83 mg KCl}$

Your solution contains 83 mg% $\text{KCl}$.

4. Promille

You know that percent by mass (%) refers to the number of grams of a substance in 100 g of a sample.

In the same way, promille (‰) refers to the number of grams of a substance in 1000 g of a sample.

You have 10 g of $\text{KCl}$ in 12 L of solution with a density of 1.5 g/mL.

The mass of your solution is

12 color(red)(cancel(color(black)("L solution"))) × (1000 color(red)(cancel(color(black)("mL solution"))))/(1 color(red)(cancel(color(black)("L solution")))) × "1 .5 g solution"/(1 color(red)(cancel(color(black)("mL solution")))) = "18 000 g solution"

Thus, you have 10 g of $\text{KCl}$ in 18 000 g of solution.

In 1000 g of solution you have

1000 color(red)(cancel(color(black)("g solution"))) × "10 g KCl"/("18 000" color(red)(cancel(color(black)("g solution")))) = "0.56 g KCl"#

Your solution contains 0.56 ‰ $\text{KCl}$.