Chapter 3, Problem 3.68PAE

(a)

Interpretation Introduction

To determine:

The initial volume (in mL) to generate 10.0 L of 0.45 M solution from 3.0 M solution.

Explanation of Solution

Molarity is defined as the number of moles of solute in 1 L of solution.

The final solution is 0.45 M, i.e. it contains 0.45 moles of solute per 1000 mL solution. Thus, for 10.0 L solution, the number of moles would be:

$n=0.45\times \frac{10.0L}{1L}=4.5\text{}moles$

To get 4.5 moles from 3.0 M initial solution, the volume of the solution required is calculated below:

The molarity of the initial solution is,

$\begin{array}{c}M=\frac{n}{VolumeofsolutioninmL}\times 1000\\ 3.0M=\frac{4.5moles}{VolumeinmL}\times 1000\\ VolumeinmL=\frac{4.5}{3}\times 1000=1500mL\end{array}$

Conclusion

Therefore, knowing the molarity and volume of the desired final solution, and the molarity of the source (initial) solution, volume of the source (initial) solution required, can be calculated.

(b)

Interpretation Introduction

To determine:

The initial volume (in mL) to generate 1.50 L of 1.0 M solution from 11.7 M solution.

Explanation of Solution

Molarity is defined as the number of moles of solute in 1 L of solution.

The final solution is 1.0 M, i.e. it contains 1.0 moles of solute per 1.0 L solution. Thus, for 1.50 L solution, the number of moles would be:

$n=1.0\times \frac{1.5L}{1L}=1.5\text{}moles$

To get 1.5 moles from 11.7 M initial solution, the volume of the solution required is calculated below:

The molarity of the initial solution is,

$\begin{array}{c}M=\frac{n}{VolumeofsolutioninmL}\times 1000\\ 11.7M=\frac{1.5moles}{VolumeinmL}\times 1000\\ VolumeinmL=\frac{1.5}{11.7}\times 1000=128.2mL\end{array}$

Conclusion

Therefore, knowing the molarity and volume of the desired final solution, and the molarity of the source (initial) solution, volume of the source (initial) solution required, can be calculated.

(c)

Interpretation Introduction

To determine:

Explanation of Solution

Molarity is defined as the number of moles of solute in 1 L of solution.

The final solution is 0.025 M, i.e. it contains 0.025 moles of solute per 1000mL solution. Thus, for 100.0mL solution, the number of moles would be:

$n=0.025\times \frac{100mL}{1000mL}=0.0025\text{}moles$

To get 0.0025 moles from 1.15 M initial solution, the volume of the solution required is calculated below:

The molarity of the initial solution is,

$\begin{array}{c}M=\frac{n}{VolumeofsolutioninmL}\times 1000\\ 1.15M=\frac{0.0025moles}{VolumeinmL}\times 1000\\ VolumeinmL=\frac{0.0025}{1.15}\times 1000=2.17mL\end{array}$

Conclusion

Therefore, knowing the molarity and volume of the desired final solution, and the molarity of the source (initial) solution, volume of the source (initial) solution required, can be calculated.

(d)

Interpretation Introduction

To determine:

Explanation of Solution

Molarity is defined as the number of moles of solute in 1 L of solution.

The final solution is $3.3\times {10}^{-5}M$, i.e. it contains $3.3\times {10}^{-5}$

moles of solute per 1000 mL solution. Thus, for 50 mL solution, the number of moles would be:

$n=3.3\times {10}^{-5}\times \frac{50mL}{1000mL}=1.65\times {10}^{-6}moles$

To get $1.65\times {10}^{-6}$

moles from 0.25 M initial solution, the volume of the solution required is calculated below:

The molarity of the initial solution is,

$\begin{array}{c}M=\frac{n}{VolumeofsolutioninmL}\times 1000\\ 0.25M=\frac{1.65\times {10}^{-6}moles}{VolumeinmL}\times 1000\\ VolumeinmL=\frac{1.65\times {10}^{-6}}{0.25}\times 1000=6.6\times {10}^{-3}mL\end{array}$