What is the total volume in milliliters ml of 12.0 M HCl that is needed to be diluted to prepare the required solution?

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.

How to Calculate Dilution

calculations are very important in Chemistry.

First we need to understand what a dilution is: 
A dilution is when you have a solution of a certain concentration and you add more solvent (remember: the substance that does the dissolving!) to decrease the concentration. If you are adding more solvent, the volume of the whole solution is going to increase as the concentration of the solution decreases. You can solve for the concentration or volume of the concentrated or dilute solution using the equation: M1V1 = M2V2, where M1 is the concentration in molarity (moles/Liters) of the concentrated solution, V2 is the volume of the concentrated solution, M2 is the concentration in molarity of the dilute solution (after more solvent has been added), and V2 is the volume of the dilute solution.

Let’s do an example! Say you have a 12 M (M = moles/L) solution of hydrochloric acid. You want to dilute this concentrated HCl with water so that you have 50 mL of a 3 M hydrochloric acid solution. How many milliliters of the 12 M HCl will you need to prepare the 3 M solution? Well, let’s go back to our dilution equation: M1V1 = M2V2. To use this equation, we need to figure out what the problem is giving us. There is a concentrated 12 Molar HCl solution (M1) and we want to end up with 50 milliliters (V2) of a 3 Molar HCl solution (M2). So, we are solving for V1: how much of the concentrated solution we will need. Plugging the numerical values into the equation we get: (12 moles/L)(V1) = (3 moles/L)(50 mL). We can then multiply 3 moles/L by 50 mL to get 150 mL * moles/L. Then we divide 150 mL * moles/L by 12 moles/L (the units moles/L will cancel out) and we get 12.5 mL! So, we need 12.5 mL of the 12 M HCl diluted with water up to 50 mL to create a 3 Molar HCl solution.

- By TutaPoint.com tutor Kaitlyn Koenig. Kaitlyn tutors in Physics, Physical Science, Chemistry, Basic Math Skills, Algebra 2, Algebra 1, Mandarin Chinese at TutaPoint.com

• Subject : Science
• Topic : Chemistry
• Posted By : Admin

1. 1M HCl: add 1mol/12M = 83 ml conc. HCl to 1L of water or 8.3ml to 100ml.
2. 2M HCl: add 2mol/12M = 167 ml conc. HCl to 1L of water or 16.7ml to 100ml.

What is 12M HCl?

12M (37% HCL) = 12 moles/L = 12 x 36.5 = 438 g/L = 438 mg/ml.

What volume of 12M HCl solution is required?

0.0104L

What is the volume of 10 m HCl?

428 mL

How many mm is 12.0 m HCl?

44.17 milliliters (mL) of a 12.0 M HCl solution is needed to make 530 mL of a 1.00 M solution.

How do you dilute 12m HCl to 3 m HCl?

So, we need 12.5 mL of the 12 M HCl diluted with water up to 50 mL to create a 3 Molar HCl solution.

What volume in milliliters of concentrated HCl 12 m is needed to make 1500 mL of a 3.5 M solution?

0.292 liters of 12 M HCl can make 1 liter of 3.5 M HCl, but the question asks for 1.5 liters. To get this, multiply 0.292 liters by 1.5, and the new result, 0.4375, represents the amount of 12 M HCl required to prepare a 1500 mL 3.5 M HCl solution.

What volume of 12M HCl is required to make 75 mL of 3.5 M HCl?

21.9 mL

What mass of KBr do you need to make 250.0 mL of a 1.50 KBr solution?

44.6 g

What is called normality?

Normality (N) is defined as the number of mole equivalents per liter of solution:normality = number of mole equivalents/1 L of solution. Like molarity, normality relates the amount of solute to the total volume of solution; however, normality is specifically used for acids and bases.

What is acid normality?

In acid-base chemistry, normality is used to express the concentration of hydronium ions (H3O+) or hydroxide ions (OH−) in a solution. Here, 1feq is an integer value. Each solute can produce one or more equivalents of reactive species when dissolved.

What does normalcy mean?

: the state or fact of being normal a return to normalcy after war.

Who first said normalcy?

Warren Harding

How do you use the word normalcy?

Normalcy sentence example

1. The sense of normalcy faded as they moved through the mall.
2. Dean said, trying to force a tone of normalcy into his strained voice.
3. Mr.
4. In an effort at normalcy , she broke the quiet as Martha dallied over the meal.

What’s another word for normalcy?

What is another word for normalcy?

Related

Your strategy here will be to use the molarity and volume of the diluted solution to determine how many moles of solute, which in your case is hydrochloric acid, #"HCl"#, it must contain..

Once you know this value, you an sue the molarity of the stock solution as a conversion factor to see how many milliliters would contain this many moles.

The underlying principle of a dilution is the fact that the number of moles of solute must remain constant. Basically, a dilution decreases the concentration of a solution by increasing its volume.

As you know, a solution's molarity tells you how many moles of solute you get per liter of solution. The diluted solution has a molarity of #"3.00 mol L"^(-1)#, which means that every liter of this solution will contain #3.00# moles of hydrochloric acid.

In your case, the #"500.-mL"# sample will contain

#500. color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "3.00 moles HCl"/(1color(red)(cancel(color(black)("L")))) = "1.5 moles HCl"#

Now use the molarity of the stock solution to determine how many milliliters would contain #1.5# moles of #"HCl"#. Since a concentration of #"12.0 mol L"^(-1)# means that you get #12.0# moles of hydrochloric acid per liter of solution, you an say that you have

#1.5 color(red)(cancel(color(black)("moles"))) * "1 L"/(12.0color(red)(cancel(color(black)("moles")))) * (10^3color(white)(a)"mL")/(1color(red)(cancel(color(black)("L")))) = color(green)(|bar(ul(color(white)(a/a)"125 mL"color(white)(a/a)|)))#

The answer is rounded to three sig figs.

ALTERNATIVE APPROACH

You can get the same result by using the equation for dilution calculations, which looks like this

#color(blue)(overbrace(c_1 xx V_1)^(color(red)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(red)("moles of solute in diluted solution"))#

Here you have

#c_1#, #V_1# - the molarity and volume of the concentrated solution
#c_2#, #V_2# - the molarity and volume of the diluted solution

Rearrange to solve for #V_1# and plug in your values to find

#c_1V_1 = c_2V_2 implies V_1 = c_2/c_1 * V_2#

#V_1 = (3.00 color(red)(cancel(color(black)("M"))))/(12.0color(red)(cancel(color(black)("M")))) * "500. mL" = color(green)(|bar(ul(color(white)(a/a)"125 mL"color(white)(a/a)|)))#