Can someone please answer this question

Question

You need to dilute a solution of 34 M hydrochloric acid. You need the dilute solution to be 6 M and need 740 mL of it. How many mL of the concentrated solution are needed?

 mL

Answer 1

Hey Janet,

In order to answer this question, we'll need to use the dilution equation: M₁V₁= M₂V₂

This equation tells us that the product of the initial concentration and initial volume will equal the product of the final concentration and final volume.

What we know:

We have been given three of these four unknowns in this problem:

Initial Concentration (M₁) = 34 mol/L *this is what we're starting with- what we want to dilute with water

Final Concentration (M₂) = 6 mol/L *this is the less-concentrated version that we're trying to create

Final Volume (V₂) = 740 mL *this is how much of the diluted solution we'd like to end up with

What we need to find out:

We've been asked to calculate how much of the initial, highly concentrated solution we should start with in order to create the 740 mL of the diluted solution:

Initial Volume (V₁) = unknown *this is what we want to find

Steps to Solve:

From here, we just need to plug in the given information into our dilution equation, and solve for our unknown variable (V₁). Here's what that looks like:

Dilution Equation: M₁V₁= M₂V₂

Plugging in given info: (34 mol/L)(V₁) = (6 mol/L)(740 mL)

Dividing both sides by M₁:

(34 mol/L)(V₁) = (6 mol/L)(740 mL)

----------------- -----------------------

(34 mol/L) (34 mol/L)

This gives us:

(6 mol/L)(740 mL)

V₁ = ----------------------- *notice that mol/L cancels out on top/bottom leaving us with mL

(34 mol/L)

V₁ = 130.588 mL

Reminder: be sure to round using sig figs for your final answer

Hope this helps!