How many moles of HCl in 25ml of 1.0 M HCl?

Answer 1

What you are wanting to do is reduce the concentration so that it is 0.24/1.5 (=0.16) of what it currently is. To get this you need to incrase the ammount of water by 1.5/0.24 (=6.25). So you will need to add (6.25*25)-25 ml of water (the -25 is because you already have 25 ml of water). So your answer is 131.25 ml The only way to get good at these is to practice, it always helps to write down exactly what you have in words and to remember that molar means the ammount in 1000 ml.

Answer 2

400 mL. (100 mL)(1.0 M HCl)=(0.25 M HCl)v v is the volume that you are trying to find. I believe the answer is 300 ml. as the question is "How much do you add . . ." Total volume is 400 ml; so must add 300ml to the original 100 ml

Answer 3

Use the formula:

M1V1 = M2V2

Where:
M1 is the molarity of the first solution (1.50)
V1 is the volume of the first solution
M2 is the molarity of the new solution
V2 is the TOTAL volume of the new solution

Grab a calculator, and subtract 138 from V2

Answer 4

This question has no meaning as written. One cannot dilute .75 milliliters (mL) to make a smaller volume of 0.25 milliliters.

I suspect that the question should read "What is the volume of water that must be added to 300 mL of .75 M HCl to dilute the solution to 0.25 M?" This is a sensible question. M refers to molarity, or moles per liter.

To solve this problem and others like it, use the equation M1V1 = M2V2.

M1 = the molarity of the initial/more concentrated solution

V1 = the volume of the initial/more concentrated solution

M2 = molarity of the second/more dilute solution

V2 = volume of the initial/more dilute solution

Plug in the known values, rearrange and solve algebraically. For the problem above, this is (M1V1/M2 = V2, or (.75 M)(300 mL)/(.25M) = 224.75 mL needed.

Answer 5

Use the formula C1xV1=C2xV2.

C1= concentration of the first solution

V1= volume of first solution

C2 and V2 are the concentration and volume of the second solution.

C1=12.0 M

V1=unknown

C2=2.5 M

V2=250 mL

Solve for V1=(C2xV2)/C1=(2.5x250)/12=52.08 mL

note: this formula will work for any concentration and volume units as long as both concentrations or volumes have the same unit measurement (ounces, liters, percent, ect).

Answer 6

You're going to dilute the 1 M HCl in water to get a less concentrated 0.25 M HCl. First determine the amount of moles in the final 10 mL 0.25 M HCl.

10 mL = .01 L × .25 moles/L = .0025 moles HCl

Now you work backwards with the concentrated solution.

1 mol/L × (volume) = .0025 moles HCl

Volume = .0025 liters = 2.5 mL

You need to prepare a 10 mL aqueous solution containing 2.5 mL of 1 M HCl.

Answer 7

There would be 0.025 moles of HCl in 25ml of 1.0 M HCl solution. This is calculated using the formula moles = molarity × volume (in liters).

Answer 8

(505ml)(0.125M HCl) = (Xml)(0.100M)

add 63.125ml

Answer 9

Molarity = moles of solute/Liters of solution ( 25 ml = 0.025 L )

1.0 M HCl = moles HCl/0.025 Liters

= 2.5 X 10^-2 moles of HCL

Answer 10

V1M1 = V2M2(x ml)(3.0 M) = (250 ml)(1.2 M)

x = 100 mls