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.
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
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
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.
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).
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.
(505ml)(0.125M HCl) = (Xml)(0.100M)
add 63.125ml
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
V1M1 = V2M2(x ml)(3.0 M) = (250 ml)(1.2 M)
x = 100 mls
The answer is 0,274 moles.
One step at a time.1/103 = 0.001 M HCl, so.....Molarity = moles of solute/Liters of solution ( 25 ml = 0.025 Liters )0.001 M HCl = X moles HCl/0.025 Liters= 2.5 X 10 - 5 moles HCl========================now, balanced eqiationNaOH + HCl --> NaCl + H2O ( all one to one )( now drive reaction towards mass NaOH )2.5 X 10 - 5 moles HCl (1 moles NaOH/1 mole HCl)(39.998 grams/1 mole NaOH)= 10 -4 grams caustic soda needed==========================
Divide by molar mass and check the units(italicalized):0.140 (g HCl) / 36.45 (g.mol-1HCl) = 3.84*10-3 mol HCl
1. First, remember definition of M (moles), M = moles of species / L. 0.33 M = 0.33 moles HCl / L 2. Then, multiple your volume by the molar concentration: 0.33 moles HCl / L x 0.70 L = 0.231 moles HCl or you can say n=CONCENTRATION multiply by VOLUME(HCl) which gives 2310 mol HCl It's helpful to carry the units with your calculations. That way you can check that numerators and denominators cancel to give you the units of your answer.
Find moles of HCl first. 1.56 grams HCl (1mole HCl/36.458 grams) = 0.0428 moles HCl Molarity = moles of solute/volume of solution Molarity = 0.0428 moles/26.8 ml = 0.00160 milli-Molarity, or more to the point; = 1.60 X 10^-6 Molarity of HCl
The answer is 0,274 moles.
One step at a time.1/103 = 0.001 M HCl, so.....Molarity = moles of solute/Liters of solution ( 25 ml = 0.025 Liters )0.001 M HCl = X moles HCl/0.025 Liters= 2.5 X 10 - 5 moles HCl========================now, balanced eqiationNaOH + HCl --> NaCl + H2O ( all one to one )( now drive reaction towards mass NaOH )2.5 X 10 - 5 moles HCl (1 moles NaOH/1 mole HCl)(39.998 grams/1 mole NaOH)= 10 -4 grams caustic soda needed==========================
Divide by molar mass and check the units(italicalized):0.140 (g HCl) / 36.45 (g.mol-1HCl) = 3.84*10-3 mol HCl
507 x (6.022 x 10^23)
1. First, remember definition of M (moles), M = moles of species / L. 0.33 M = 0.33 moles HCl / L 2. Then, multiple your volume by the molar concentration: 0.33 moles HCl / L x 0.70 L = 0.231 moles HCl or you can say n=CONCENTRATION multiply by VOLUME(HCl) which gives 2310 mol HCl It's helpful to carry the units with your calculations. That way you can check that numerators and denominators cancel to give you the units of your answer.
The limiting reagent in a reaction is the reactant that runs out first. For example, if you are reacting 10 moles of HCl and 5 moles of NaOH, you will get 5 moles of H20, 5 moles of NaCl, and 5 moles of HCl, because the remaining HCl had nothing to react with. Therefore, the NaOH is the limiting reagent.
Find moles of HCl first. 1.56 grams HCl (1mole HCl/36.458 grams) = 0.0428 moles HCl Molarity = moles of solute/volume of solution Molarity = 0.0428 moles/26.8 ml = 0.00160 milli-Molarity, or more to the point; = 1.60 X 10^-6 Molarity of HCl
First of all, it is HCl solution, more properly hydrochloric acid (that is HC l ). It is composed of hydrogen (H) and chlorine (Cl). The answer depends on the volume of the acid and the concentration of the alkali. Here is a sample calculation with some random values for the variables you have not given. If the HCl is neutralized by 25.0 ml of 0.500 M NaOH, then the number of moles of NaOH equals the number of moles of HCl. 25 ml is equal to 0.025 liters, and since molarity is moles per liter we have: Moles of NaOH = 0.0250 Liters * 0.500 moles/liter = 0.0125 moles Moles of NaOH = moles of HCl = 0.0125 moles If there are 0.0125 moles HCl in 45.0 ml (or 0.045 L), then the molarity of the HCl is: 0.0125 moles ÷ 0.0450 L = 0.278 moles/L = 0.278 M HCl
To determine the number of moles of ions present in a known volume of solution, follow this example:HCl dissociates completely in water into H+ and Cl-, because this is a strong acid, and only strong acids, bases, and ionic compounds have the ability to dissociate completely.This means one equivalent of HCl will generate one equivalent of H+ and Cl- ions; the same number of moles of HCl will generate the same number of moles for H+ and Cl-HCl --> H+ + Cl-Now determine the number of moles in the volume of your solution. Remember that 1M is another way to say 1 mole/L.(2moles HCl/ 1L) x (1L) = 2 moles HClSince the equation states that 1 equivalent of HCl is 1 H+, the final answer is:(2moles HCl/ 1L) x (1L) x (1 mole H+/1mole HCl) = 2 moles H+
The answer to the conversion is that 35.0 grams of hydrochloride (HCL) equals 0.76 moles. The conversion rate is 35.0 grams divided by 46 gram per mole. A mole is the molecular weight of a substance.
What you are doing here is titration. You know you have a solution of HCl, but you do not know how much HCl is in it. For this you use something that can react with HCl (NaOH) and use an indicator to tell you when the reaction is complete. The reaction is pretty simple: HCl + NaOH --> H2O + NaCl You can see here that NaOH and HCl have a 1:1 mol relationship. So, lets find out how many moles of NaOH you used up with concentration = moles/volume 0.10 M NaOH = moles NaOH/ 0.0197 L NaOH solution Remember that M is in moles/L moles NaOH = 0.00197 moles Since you have a 1:1 relationship of NaOH with HCl, the 0.00197 mol applies to HCl as well. The next question works the same way, but backwards. Try doing it yourself if you understood the first part before reading my answer. Find out how many moles of HCl you have so you can find out how much moles of NaOH you need for the neutralization. 0.050 M HCl = n HCl / 0.020 L HCl soln n HCl = 0.001 mol HCl Remeber the 1:1 relationship, which gives you that n NaOH = 0.001 mol. Now all you need is the volume. 0.1 M NaOH = 0.001 mol NaOH/ Volume soln V = 0.01 L = 10 mL
That is approximately 1/10 of a cup