Chapter U4.15, Problem 4E

Interpretation Introduction

Interpretation: The given table needs to be completed for glucose.

Concept Introduction: The molarity of solution is defined as number of moles of solute in 1 L of the solution.

  $M=\frac{n}{V\left(L\right)}$

Here, n is number of moles and V is volume of solution in L.

The number of moles is calculated from mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Explanation of Solution

The given table is as follows:

According to the given table:

(1)

In $\text{2 L}$ solution there is $\text{180g}$ glucose.

Thus, in $\text{1 L}$ solution there is $\frac{\text{180}}{\text{2}}\text{g=90g}$ glucose.

Molar mass of glucose is 180 g.

The number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Thus,

$\text{90g}$ glucose $\text{=}\frac{\text{90}}{\text{180}}\text{=0}\text{.5}$ mol.

Therefore, molarity is 0.50 M

(2)

Now, $\text{0}\text{.40}$ mol of glucose is present in $\text{1 L}$ .

Thus,

$\text{1}$ mol of glucose is present in $\frac{\text{1}}{0.40}$ L.

Also,

$\text{0}\text{.10}$ mol of glucose is present in $\frac{\text{1}\times \text{0}\text{.10}}{0.40}$ L= $\text{0}\text{.25L}$

Mass of $0.10$ mol of glucose $\text{=0}\text{.10×180g=18g}$ .

(3)

$\text{0}\text{.10}$ mol of glucose is present in $\text{1 L}$ .

Thus,

$0.03$ mol of glucose is present in $\frac{\text{1×0}\text{.03}}{\text{0}\text{.10}}\text{L}$ = $\text{0}\text{.3L}$ .

Mass of $0.03$ mol of glucose $\text{=0}\text{.03×180g=5}\text{.4g}$

(4)

$\text{180g}$ glucose is 1 mol.

Thus,

$\text{1g}$ glucose is $\frac{\text{1}}{180}$ mol.

Or,

$\text{0}\text{.180g}$ glucose $=\frac{\text{1}\times \text{0}\text{.180}}{180}$ mol $=0.001\text{ mol}$ .

Thus,

$\text{0}\text{.1L}$ solution contains $0.001$ mol of glucose.

And,

1L solution contains $\text{=}\frac{0.001}{0.1}$ mol of glucose $=0.01$ mol of glucose

Thus, molarity of glucose is 0.01 M.

Therefore, the complete table will be as follows:

Molarity(moles/liter)	Volume(liters)	Number(moles of solute)	Molar mass of solute(grams/mole)	Mass of solute(grams)
0.50 M	2.0 L	1 mol	180 g/mol	180 g
0.40 M	0.25 L	0.10 mol	180 g/mol	18 g
0.10 M	0.3 L	0.03 mol	180 g/mol	5.4 g
0.01	100 mL=0.1 L	0.001 mol	180 g/mol	0.180 g