Chapter 7.3, Problem 49E

a.

To determine

To calculate: the number of liters each solution will satisfy when we use 2 liters of the 50% solution.

Answer to Problem 49E

1l of 10% solution

7l of 20% solution

2l of 50% solution

Explanation of Solution

Given:

A chemist needs 10 liters of 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%.

Calculation:

Let $x$ be amount of 10% solution, y be amount of 20% solution and z be amount of 50% solution.

So, $x+y+z=10\cdots \left(i\right)$

According to question $0.1x+0.2y+0.5z=0.25\times 10\cdots \left(ii\right)$

Here value of z is 2

By putting value of y in above equations we get

  $\begin{array}{l}x+y=8\cdots \left(iii\right)\\ 0.1x+0.2y=1.5\cdots \left(iv\right)\end{array}$

Now multiplying equation $\left(iv\right)$ with 10 we get

  $x+2y=15\cdots \left(v\right)$

Subtracting equation $\left(iii\right)$ from $\left(v\right)$ , we get

  $y=7$

Put $y=7$ in equation $\left(iii\right)$ , we get

  $x=1$

and $z=2$ is given

So,

1l of 10% solution

7l of 20% solution

2l of 50% solution

Conclusion:

1l of 10% solution

7l of 20% solution

2l of 50% solution

b.

To determine

To calculate: the number of liters each solution will satisfy when we use as little as possible of the 50% solution.

Answer to Problem 49E

0l of 10% solution

8.33l of 20% solution

1.67l of 50% solution

Explanation of Solution

Given:

A chemist needs 10 liters of 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%.

Calculation:

Let $x$ be amount of 10% solution, y be amount of 20% solution and z be amount of 50% solution.

So, $x+y+z=10\cdots \left(i\right)$

According to question $0.1x+0.2y+0.5z=0.25\times 10\cdots \left(ii\right)$

To use as little as possible of 50% solution means mix it with 20% solution only

So here value of x is 0

By putting value of x in above equations we get

  $\begin{array}{l}y+z=10\cdots \left(iii\right)\\ 0.2y+0.5z=2.5\cdots \left(iv\right)\end{array}$

Now multiplying equation $(iv)$ with 10 we get

  $2y+5z=25\cdots \left(v\right)$

Multiplying equation $\left(iii\right)$ with 2

  $2y+2z=20\cdots \left(vi\right)$

Subtracting equation $\left(vi\right)$ from $\left(v\right)$ , we get

  $\begin{array}{l}3z=5\\ z=1.67\end{array}$

Put $z=1.67$ in equation $\left(iii\right)$ , we get

  $y=8.33$

So

0l of 10% solution

8.33l of 20% solution

1.67l of 50% solution

Conclusion:

0l of 10% solution

8.33l of 20% solution

1.67l of 50% solution

c.

To determine

To calculate: the number of liters each solution will satisfy when we use as much as possible of the 50% solution.

Answer to Problem 49E

6.25l of 10% solution

0l of 20% solution

3.75l of 50% solution

Explanation of Solution

Given:

A chemist needs 10 liters of 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%.

Calculation:

Let x be amount of 10% solution, y be amount of 20% solution and z be amount of 50% solution.

So, $x+y+z=10\cdots (i)$

According to question $0.1x+0.2y+0.5z=0.25\times \mathrm{10...}(ii)$

To use as much as possible of 50% solution means mix it with 10% solution only.

So here value of y is 0

By putting value of y in above equations we get

  $\begin{array}{l}x+z=10\cdots \left(iii\right)\\ 0.1x+0.5z=2.5\cdots \left(iv\right)\end{array}$

Now multiplying equation $(iv)$ with 10 we get

  $x+5z=25\cdots \left(v\right)$

Subtracting equation $\left(iii\right)$ from $\left(v\right)$ , we get

  $\begin{array}{l}4z=15\\ z=3.75\end{array}$

Putting $z=3.75$ in equation $\left(iii\right)$ , we get

  $\begin{array}{l}x+3.75=10\\ x=6.25\end{array}$

So

6.25l of 10% solution

0 of 20% solution

3.25l of 50% solution

Conclusion:

6.25l of 10% solution

0 of 20% solution

3.25l of 50% solution