# Question #d232c

##### 2 Answers

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

The first thing to do here is to figure out how many **moles** of fructose must be dissolved in water in order to get

Use the fact that

#color(blue)(ul(color(black)("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"))))#

#=" 0.0140 moles fructose"#

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

#0.0140 color(red)(cancel(color(black)("moles fructose"))) * "180 g"/(1color(red)(cancel(color(black)("mole fructose")))) = color(darkgreen)(ul(color(black)("2.5 g"))) -># two sig figs

For part **(b)**, you know that you have a

Notice that the concentration of the *diluted solution* is **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) -># thedilution factor

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

#"DF" = V_"diluted"/V_"stock" implies V_"stock" = V_"diluted"/"DF"#

Keep in mind that the **dilution factor**, *compared with* the diluted solution.

You will thus have

#V_"stock" = "375 cm"^3/color(blue)(6) = "63 cm"^3 -># two sig figs

So, in order to prepare this solution, you would start with *enough water* to get the total volume of the solution to

For part **(c))**, I'll assume that you must express the concentration of the *milligrams per* *cubic centimeters*,

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**

#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

#color(blue)(ul(color(black)("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.

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

**1. Molarity**

∴

**2. Millimolarity**

Here, we must use the conversion

**3. Mg percentage**

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

You have 10 g of

∴ In 100 mL of solution, you have

Your solution contains 83 mg%

**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

The mass of your solution is

Thus, you have 10 g of

In 1000 g of solution you have

Your solution contains 0.56 ‰