A nurse needs to treat a patient with an infection, using a 0.25% saline solution. He prepared the saline solution by diluting 50.0mL of a stock saline solution with a concentration of 3.0%. What...

A nurse needs to treat a patient with an infection, using a 0.25% saline solution. He prepared the saline solution by diluting 50.0mL of a stock saline solution with a concentration of 3.0%. What volume of dilute 0.25% saline did the nurse prepare.

Asked on by daegnam

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steveschoen's profile pic

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

It seems like there is something missing here.  So, we will do what we can.

We can set up a system of equations here.  We need two variables.  I am assuming it was water we diluted the saline with.  We don't know how much water we used; that's x.  Also, we don't know how much of the 0.25% solution we ended up with; that's y.  But, we do know when we add the water to the 50 ml solution, we get the 0.25% solution.  So:

50 + x = y

Then, we can use a formula for the second equation.  For percent formula problems, trying to make a formula out of it:

Amount#1 * %#1  +  Amount#2 * %#2  +  Amount#3 * %#3 + . . . = Amount(total) * %total

So, filling in the numbers and variables we have:

50*0.03 + x*0 = y*0.0025

Simplifying each equation:

50+x = y

1.5 = 0.0025y

Solving the second equation for y:

y = 600 ml

And, that is the what we are looking for, how much 0.25% saline solution the nurse prepared.  X stood for how much water was added to achieve this.  Substituting the y back into the first equation for it, we would get:

50+x = 600

x = 550 ml

elianagerzon's profile pic

elianagerzon | eNotes Newbie

Posted on

This problem can be answered by the formula C1V1=C2V2. For this formula you have two concentrations and two volumes. The first concentration (C1)is the concentration of the stock solution. The stock solution is the source of solution that the nurse will dilute. The V1 is the volume of the stock solution. From V1 the volume will be increase to V2 to attain the C2 concentration. It means that C2 is the final concentration and V2 is the final volume.

C1V1=C2V2

(3.0%)(50mL)=(0.25%)(V2)

V2=600 mL

The answer for this problem is 600 mL

You will not need to subtract 50mL from 600 mL because the problem is asking for the volume of the dilute 0.25% saline solution.

But if the question asks for the volume added to dilute the solution, you will need to subtract 50 from 600.

ayl0124's profile pic

ayl0124 | Student, Grade 12 | (Level 1) Valedictorian

Posted on

The dilution equation is:

`C_1V_1 = C_2V_2`

"C" stands for concentration, while "V" stands for volume.

Your initial concentration is 3.0% and your initial volume is 50 mL. Your new concentration is 0.25%.` `

`(3.0%)(50) = (0.25%)(V_2)`

Solve for your unknown variable.

`V_2 = 600 "mL"`

The total volume of the new solution is 600 mL. Since the lowest sig fig is two, your answer should have two sig figs. Thus, your answer is: 

`V_2 = 6.0 xx 10^2 "mL"`

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