Chapter U4.81, Problem 8E

(a)

Interpretation Introduction

Interpretation:

Number of moles of solute must be found out in 50 L of 0.10 M NaCl.

Concept Introduction :

The number of moles of solute can be calculated from molarity and volume as follows:

  $n=M\times V$

Here, M is molarity and V is volume.

Answer to Problem 8E

Moles of solute in 50 L of 0.10 M NaCl is 5.

Explanation of Solution

Molarity(M)=0.10

Volume (V) of solution is 50.0 L

hus, moles of NaCl in 50.0 L of 0.10 M NaCl can be calculated as follows:

  $\begin{array}{l}n=M\times V\\ =\left(0.1\text{ mol/L}\right)\left(50\text{ L}\right)\\ =5\text{ mol}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Number of moles of solute must be found out in 0.25 L of 3. 0 M glucose.

Concept Introduction :

The number of moles of solute can be calculated from molarity and volume as follows:

  $n=M\times V$

Here, M is molarity and V is volume.

Answer to Problem 8E

Moles of solute in 0.25 L of 3.0 M Glucose is 0.75 mol.

Explanation of Solution

Molarity(M)=3.0

Volume (V) of solution is 0.25 L

Thus, moles of glucose in 0.25 L of 3. 0 M Glucose will be:

  $\begin{array}{l}n=M\times V\\ =\left(\text{3 mol/L}\right)\left(0.25\text{ L}\right)\\ =0.75\text{ mol}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Number of moles of solute must be found out in 35 mL of 12. 0 M HCl.

Concept Introduction :

The number of moles of solute can be calculated from molarity and volume as follows:

  $n=M\times V$

Here, M is molarity and V is volume.

Answer to Problem 8E

Moles of solute in 35 mL of 12. 0 M HCl is 0.42 mol.

Explanation of Solution

Molarity(M)=12.0 M

Volume (V) of solution is 35 mL=35/1000 L=0.035 L

hus, moles of solute in 35 mL of 12. 0 M HCl is

  $\begin{array}{l}n=M\times V\\ =\left(\text{12 mol/L}\right)\left(0.035\text{ L}\right)\\ =0.42\text{ mol}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Number of moles of solute must be found out in 300 mL of 0.025 M NaOH.

Concept Introduction :

The number of moles of solute can be calculated from molarity and volume as follows:

  $n=M\times V$

Here, M is molarity and V is volume.

Answer to Problem 8E

Moles of solute in 300 mL of 0.025 M NaOH is 0.0075.

Explanation of Solution

Molarity(M)=0.025.

Volume (V) of solution is 300 mL=300/1000 L=0.3 L

Thus, moles of solute in 300 mL of 0.025 M NaOH will be:

  $\begin{array}{l}n=M\times V\\ =\left(\text{0}\text{.025 mol/L}\right)\left(0.3\text{ L}\right)\\ =0.0075\text{ mol}\end{array}$