Chapter U4.15, Problem 5E

(a)

Interpretation Introduction

Interpretation:

The volume of 0.10 M NaCl containing 58.44 g NaCl needs to be determined.

Concept Introduction :

Molecular weight of NaCl is 58.44 g/mol.

Molarity of NaCl is 0.10 means there is 0.10 mol NaCl in 1 L solution.

Answer to Problem 5E

10 L of 0.10 M NaCl would contain 58.44 g NaCl.

Explanation of Solution

Molecular weight of NaCl is 55.44 g/mol.

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

  $m=n\times M$

I L must contain 0.10 mole NaCl $=0.10\text{ mol}\times 58.44g/mol=5.844g$

5.844 g NaCl would be in 1L solution.

So, 58.44 g NaCl is would be in 10 L solution.

(b)

Interpretation Introduction

Interpretation:

The volume of 3.0 M NaCl containing 58.44 g NaCl needs to be calculated.

Concept Introduction :

Molecular weight of NaCl is 58.44 g/mol.

Molarity of NaCl is 3.0 means there is 3.0 mol NaCl in 1 L solution.

Answer to Problem 5E

0.33 L of 3.0 M NaCl would contain 58.44 g NaCl.

Explanation of Solution

Molecular weight of NaCl is 55.44 g/mol.

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

  $m=n\times M$

I L must contains 3.0 mole NaCl which is equal to $3.0\text{ mol}\times 58.44g/mol=175.32\text{ g}$

Since, 175.32 g is in 1 L solution, thus, 58.44 g will be in:

  $V=\frac{58.44\text{ g}}{175.32\text{ g/L}}=0.33\text{ L}$

So 58.44 g NaCl is would be in 0.33 L solution.

(c)

Interpretation Introduction

Interpretation:

The volume of 5.5 M NaCl containing 58.44 g NaCl needs to be determined.

Concept Introduction :

Molecular weight of NaCl is 58.44 g.

Molarity of NaCl is 5.5 means there is 5.5 mol NaCl in 1 L solution.

Answer to Problem 5E

0.182 L of 5.5 M NaCl would contain 58.44 g NaCl.

Explanation of Solution

Molecular weight of NaCl is 55.44 g/mol.

The mass from 5.5 mol can be calculated as follows:

  $m=n\times M=\left(5.5\text{ mol}\right)\left(58.44\text{ g/mol}\right)=321.42\text{ g}$

Thus, 1 L of solution contains 321.42 g of NaCl.

The volume of solution containing 58.44 g of NaCl will be:

  $V=\frac{58.44\text{ g}}{321.42\text{ g/L}}=0.182\text{ L}$