Chapter 3.7, Problem 41E

a.

To determine

To write: the function for the given situation

Answer to Problem 41E

  $C\left(t\right)=\frac{480+3t}{40+t}$

Explanation of Solution

Given information:

Volume of solution required for dilution = 40 liters

Concentration of solution that is diluted = 12 molar

Concentration of solution added for dilution = 3 molar

The concentration of mixture as a function number of liters t of 3 molar solution added

Calculation:

Let concentration of mixture be C and molar concentration be t. The formula to evaluate concentration of a solution (in molar) terms is as follows:

  $\begin{array}{l}C=\frac{n}{t}\\ or\\ n=Ct\end{array}$

Here n is number of moles and $t$ is the volume of solution in liters.

The function for the given situation will be evaluated as follows:

  $\begin{array}{c}C\left(t\right)=\frac{\left(n_{dilute}\right)+\left(n_{added}\right)}{t_{diluted}+t_{added}}\\ =\frac{\left(C_{dilute}\right)\left(t_{diluted}\right)+\left(C_{added}\right)\left(t_{added}\right)}{t_{diluted}+t_{added}}\\ =\frac{\left(40\right)\left(12\right)+\left(3\right)\left(t\right)}{40+t}\\ =\frac{480+3t}{40+t}\end{array}$

Thus, the function is $C\left(t\right)=\frac{480+3t}{40+t}$ .

b.

To determine

To find: the volume (in liters).

Answer to Problem 41E

11.43 liters

Explanation of Solution

Given information:

Concentration that is being added = 3 molar

New concentration = 10 molar

The concentration function of 3 molar added solution evaluated is as follows:

  $C\left(t\right)=\frac{480+3t}{40+t}$

Calculation:

Put $C\left(t\right)=10$ in the given function and evaluate t as follows:

  $\begin{array}{c}10=\frac{480+3t}{40+t}\\ 10\left(40+t\right)=480+3t\\ 400+10t=480+3t\\ 10t-3t=480-400\\ 7t=80\\ t=\frac{80}{7}\\ =11.43liters\end{array}$

Thus, the required volume is 11.43 liters.