Question

a nurse must administer an aqueous Nacl saline solution to a patient intravenously. he must prepare 225ml of a dilute solution with a concentration of 0.154 mol/L only has access to a concentrated solution with a concentration of 0.369mol/La) What volume of the concentrated solution will he need to prepare his dilute solution?b) what is the mass of Nacl contained in the prepared solution?c) The volume of concentrated solution used contains the same amount of solute as the volume of solution prepared. true or false? explain briefly using a few sentences or a calculation.

Answer 1

In order to calculate volume of the concentrated solution we have to calculate the dilution factor.

`\begin{gathered} Dilution\text{ }Factor=\frac{initial\text{ }concentration}{final\text{ }concentration}=\frac{final\text{ }volume}{initial\text{ }volume} \\ \\ \end{gathered}`

A) Dilution factor = (0.369 mol/L) / (0.154 mol/L)

D.F = 2.396

Initial volume = final volume/ Dilution factor

Initial volume = 225 mL/2.396

Initial volume = 93.907 mL

B) In order to find the mass of NaCl in the prepared solution we must first find the moles present in the solution.

moles = (0.154 mol x 225 mL) / 1000 mL

moles = 0.03465 mol

mass= molar mass of NaCl x moles

mass = 58.440 g/mol x 0.03465 mol

mass= 2.025 g

C) To determine if the statement is true or false we can calculate the moles in the initial concentration

moles = (0.369 mol x 93.907 mL) / 1000 mL

moles = 0.03465 mol

It is true that the volume of concentrated solution used contains the same amount of solute as the volume of solution prepared.