How many ml of 50 percent dextrose solution and how many ml of water are needed to prepare 100ml of 15 percent dextrose solution?
30 ml 50% dextrose + 70 ml water
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How many grams of calcium chloride would you add to water for a totoal volume of 500 ml to make a 5 percent solution?
5% of 100 is 5. Therefore; 5% of 500 is 25g.. 5 x 5 = 25 g
How many grams of calcium chloride would you add to water for a total volume of a 100 ml to make a 5 percent solution?
Take 5 grams of calcium chloride and dissolve it in 100ml of solution to get a 5% solution of calcium chloride. The standard way to make a weight-volume solution is to take grams of the dry substance in 100ml of volume.
Fiver percent dextrose is considered an isotonic solution. However,once it enters the body, it becomes hypotonic. The osmolality ofbody fluids and isotonic solutions are the same.
a 2.5% solution has 25mg per ml, therefore a 20% solution has 200mg per ml.
D5W or Dextrose 5% Water is indicated for:. Rehydration . Provides calories for some metabolic needs (100ml provides 5g dextrose) -- useful in traumas when there's a shock related calorie burn. . By adding carbohydrates, reduces metabolism of proteins for calorie uptake. . May produce diuresis. …. Used for keeping an IV catheter open, and providing a means of piggybacking meds into the system IV. . Note that emergency services more typically use normal saline or ringers lactate.. (MORE)
How many milliliters of water would you add to 100 ml of 1.0 m hcl to prepare a final solution of 0.25 m hcl?
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
90. mL NaOH, remember there are two sig figs in this problem not one, hence the decimal point after the 90.
To make a 10 percent solution of salt in water how many grams of salt should you dissolve into 100 ml of water?
10% salt solution is the equivalent of adding 100gr salt in a 900 ml (1000ml -100ml) of water. you now have one liter of a 10% solution. . Similarly 10 gr of salt in 90 ml water will give you 100 ml of a 10% salt solution. . However if you must use 100 ml of water you need to add 11 gr of salt to …get a a 10% salt solution. . Hope that helps. . TA (MORE)
How many ml of 25percent dextrose and needed to prepare 500 ml of 40percent dextrose if you are to make 40percent dextrose from 25percent dextrose and 60percent dextrose?
My calulation says 286ml of 25% and 214ml of 60%. 286 x .25= 71.50. 214 x .60= 128.40. 250 x .20= 50.00. 250 x .60= 150.00
What is the pH of a solution prepared by diluting 3.0 ml of 2.5 M HCl to a final volume of 100mL with water?
.003 L HCL * 2.5 moles HCL/1L = .0075 moles HCL .0075 moles HCL/ .1 L HCL = .075 M HCL = .075 M H3O+ pH = -log[H3O+] pH = -log[.075] = 1.12 this of course is not adjusted for activities of the ions in solution, but it would be very close to this value.
If an iv solution contains 5 percent wv of dextrose how many mg of dextrose is added by intravenous injection of 475 ml of this solution?
23.75 mg of Dextrose is added by intravenous injection of 475 ml of this solution .. wrong! . 5% WV means 5 grams/100ml ; therefore 23.75 grams/475ml; since there's 1000mg/gram . the number of mg dextrose added by IV injection of 475ml = 23750mg !
D5W or Dextrose 5% in an IV bag is clear. There should be no discoloration.. If you break open the bag and feel the solution it is sticky and runs like thin syrup.
How much pure acid should be added to 50 ml of a 10 percent acid solution to give a 50 percent acid solution?
A 50 ml solution that is 10% acid will consist of 5 ml of acid (10% of the volume) and 45 ml of water (90% of the volume). You're not adding any water, but you want to add enough acid to make a solution that is 50% acid and 50% water. You will need to have a total of 45 ml of acid in the mixture to …make it a 50/50 solution, since the amount of water is also 45 ml. You have 5 ml in there already, so you would need to add 40 ml of acid. That would make a total 90 ml solution that is 50% water (45 ml) and 50% acid (45 ml). (MORE)
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
take the formula weight for ch3oh...roughly 36multiply by 2, and then multiply that by .15 since molarity is in liters and not in ml like in the problem. answer 10.8 grams ch3oh
Dextrose is a synonym of D-glucose (also known as grape sugar, corn sugar, and when it's present in blood, blood sugar).In 2013, Dextrose 5 percent in lactated Ringer's injection wasrecalled. This recall stemmed from allegations of the producthaving mold in it.
So you have to have a milligram for every millilitre.But you have 100ml. Therefore you have to multiply 1mg by 100, toget 100mg. Weigh out 100mg or 0.1g and dissolve it to 100ml.
How many liters of a 10 percent alcohol solution must be mixed with 90 liters of a 90 percent solution to get a 50 percent solution?
Approximately 219 Liters. My math: 0.5 = (81 + 0.1 x) / (9 + 0.9 x) 0.5 (9 + 0.9 x) = (81 + 0.1 x) / (9 + 0.9 x) * (9 + 0.9 x) 4.5 + 0.45x = 81 + 0.1 x 4.5 + 0.45 x - 4.5 = 81 + 0.1 x -4.5 0.45 x = 76.5 + 0.1 x 0.45 x - 0.1 x = 76.5 + 0.1 x - 0.1 x 0.35 x = 76.5 0.35 x / 0.35 = 76.5 / 0.35 … x = 218.57143 (MORE)
M = moles/liter 100 ml = 0.1 liter so 0.01 mole NaOH / 0.1 liter = .1 M NaOH you can find how many grams of NaOH in .01 moles by multiplying .01 by the atomic weight of a mole of NaOH, which you can find by adding up the atomic weight of Na, and O, and H.
Dextrose solution is a medication in an IV form used to supply water and calories to the body. It is also used as a mixing solution (diluent) for other IV medications
Find the amount of 14 percent acid solution that should be added to 6 percent acid solution to obtain 50 mL of a 12 percent solution?
Write eqn for weight; x = 14% sol & y = 6% sol which is x + y = 50. Write eqn for solution: .14x + .06y = .12*50 = 6; 2 equations 2 unknowns. Solve: Mult 1st eqn by -.14; now -.14x-.14y = -.14*50 or -7 -.14x-.14y=-7 .14x+.06y=6 0-.08y=-1 or y=12.5 or 12.5mL of 6% solution x+12.5=50 or x=37.5 or 37.5… mL of 14% solution (MORE)
First you have to find the molar mass of NaOH which is,23+16+1=40g/mol. Then you take 0.450M X 200mL=90 mM. which can be converted to .090moles just move the decimal three decimal places. Then you take youmolar mass which is 40g/mol X .090moles= 3.6grams. (The molescancelled themselves out and your …left with grams.) Pretty easyonce you practice more problems. (MORE)
Dissolve slowly 50 g NaOH in 100 mL water; advertisement: sodium hydroxide solution is dangerous !
How many liters of a 10 percent alcohol solution must be mixed with 80 liters of a 80 percent solution to get a 50 percent solution?
The following solution is just an approximation, in practical circumstances sufficient to use. The addendum explains why. . x = number of litres (10%) to be added. You have 64 litres of alcohol in your original 80 litres. . 64 + x/10 = (80 + x)/2 . 640 + x = 400 + 5x . 4x = 240 . x = 60 and a…lcohol content (10%) will be 6. . so you will have a total of 140 litres of which 64 + 6 will be alcohol. . Added note to deal with the so called ' dilution contraction ' of total volume If it were % by MASS ( %m/m), it's quite easy to do (based on the 'Mass Conservation Law). You calculate with mass (kg) and mass-% (%m/m) i.s.o. volume (L) and vol% (%v/v). However if the meaning was: % by Volume ( %v/v) then calculation appears to become quite complicated, but not impossible if you know at least the density values of all solutions (original 80%v/v or 10%v/v and final 50%v/v). DO NOT use: (orig. volume) + (added volume) = final volume, as done above, if exact figures are necessary. It's only a rule of thump, an approximation. This is because fluids can contract on mixing at dilution. There is no rule such as: conservation of volume. Your case: 60 L + 80 L (is not equal but) < 140 L final solution. (MORE)
M V = M' V' 1M x 2L = 14.5 M x V' V' = 138 mL
50 percent dextrose solution and you need a 1 percent dextrose soultion. How do you figure that out?
If the percents given are by weight or mass, this is very straightforward: The ratio between the desired percentage and the initial percentage is 1/50. Therefore, a given mass of initial solution must be diluted to 50 times its original mass to obtain the desired lower concentration, or in other wor…ds, 49 parts of diluent must be mixed with each part of initial solution. If the percents involve volume measurements, it would be necessary to take into account and change in density occasioned by the dilution. (MORE)
Molarity = moles of solute/volume of solution 1.50 M = X Moles/125ml = 187.5 millimoles, which is 0.1875 moles The molar mass of CaCl2 = 110.98 grams 0.1875 moles CaCl2 (110.98g/1mol) = 20.8 grams needed
How many 150 mg clindamycin capsules are needed to prepare 60 ml of a 1 percent clindamycin solution?
1% 1g in 100ml = 1000mg in 100ml in 60ml ? 100x = 1000 x 60 100x = 60000 x = 60000/100 = 600 150mg ? 600/150 = 4 answer = 4
How many mL of the 0.5 M NaCl stock solution is needed to prepare 500mL of 0.25 M NaCl solution How much water is needed to dilute the stock solution?
M1V1=M2V2 .5M x ?mL = .25M X 500mL Rearrange it and you get ?mL = (.25M x 500mL) / .5 Which equals 250mL needed of the .5M NaCl.
Molarity = moles of solute/volume of solution 0.450 M = m/200ml = 90 millimoles, or, what we need; 0.09 moles 0.09 moles NaOH (39.998 grams NaOH/1 mole NaOH) = 3.60 grams of NaOH needed
How many ml of a 20 percent solution should be added to 50 ml of a 40 percent solution to obtain a 25 percent solution?
150 ml Let x be the volume of 20% solution to add. The amount of "stuff" dissolved in the 20% solution plus the amount in the 40% will equal the amount of "stuff" dissolved in the 25% solution. (0.20)(X) + (0.40)(50 ml) = (0.25)(X+50ml) 0.2X + 20 = 0.25X + 12.5 20 - 12.5 = 0.25X - 0.20X …7.5 = 0.05X X = 7.5/0.05 = 150 ml (MORE)
If you dissolve 5 grams of salt in 100 ml of water what is the concentration of the solution in percent?
Supposed 100 ml water has a mass of 99.8 grams (its density at 20 o C being 998 g/dmÂ³): Then: 100% * (5 g) / [99.8 g + 5 g] = 4.77 % salt
we need 0.8gm NaoH and dissolved in 10 ml of water to make 2N solution of NaoH .
How many milliliters shoul be added to 100ml of 10 percent stock solution of sodium chloride to prepare a 0.9 percent solution of sodium chloride?
You need to use the equation C1 x Q1 = C2 x Q2. so you just plug in your knowns: 10% x 100 = 0.9% x Q2 so you get 1,000 = 0.9% x Q2 and finally 1,111.1 mL of sodium chloride
How much of a 37 percent formaldehyde solution should be put into 900 ml of water to make a 10 percent formaledhyde solution?
If the percentages specified are by mass and the 900 ml of water are assumed to be at standard temperature and pressure, this problem can be solved as follows: The 900 ml of water will have a mass of 900 grams. Call the unknown mass of the 37 percent solution to be added m. this will contain 0.37m g…rams of formaldehyde and 0.63m grams of water, and 0.37m/(900 + m) = 0.10. Then, multiplying both sides by (m + 900), 0.37m = 0.10m + 90, or 0.27m = 90, or m = 3.3 X 10 2 grams of the 37 % solution to be added, to the justified number of significant digits. (MORE)
Molarity = moles of solute/Liters of solution ( 300.0 ml = 0.300 liters ) ( Ca(OH)2 is correct formulation ) 0.115 M Ca(OH)2 = moles Ca(OH)2/0.300 Liters = 0.0345 moles Ca(OH)2 (74.096 grams/1 mole Ca(OH)2) = 2.56 grams of Ca(OH)2 needed
add 25ml more of solution x * 20 = 100 * 25 x = 25
Molarity = moles of solute/Liters of solution ( 50 ml = 0.05 Liters ) 12 M HCl = moles HCl/0.05 Liters = 0.60 moles HCl
You should NEVER rely on animal (or human) health answers on here ! Your most accurate and qualified answer will only come from a certified veterinary practice !
How many ounces of pure water must be added to 50 ounces at a 15 percent saline solution to make a saline solution that is 10 percent salt?
Add 25 oz of pure water Let X be the volume of pure water to add. Then the total amount of pure water you add plus the total amount of water in the 15% solution will equal the total amount of water in the final 10% solution. So, X + 50*(1-0.15) = (50 + X) * (1-0.10) X + 50*(0.85) = (50 + X)*(0.9…0) X + 42.5 = 45 + 0.90X 1X - 0.90X = 45 - 42.5 0.10X = 2.5 X = 25 (MORE)
Let us find moles first. Molarity = moles of solute/Liters of solution ( 750 ml = 0.750 Liters ) 0.375 M Na 2 SO 4 = moles Na 2 SO 4 /0.750 Liters = 0.28125 moles Na 2 SO 4 === 0.28125 moles Na 2 SO 4 (142.05 grams/1 mole Na 2 SO 4 ) = 39.95 grams Na 2 SO 4 needed -------------…--------------------------you do significant figures! (MORE)
(X gal)(0.15) = (1 gal)(0.025) 0.15X = 0.025 X = 0.17 gallons (4 quarts/1 gal.)(1 liter/1.06 quarts)(1000 ml/1 liter) = 642 milliliters ====
How many grams of the compound trinitrotoluene are needed to prepare 250 ml of a 0.100 M solution of the compound?
C 7 H 5 N 3 O 6 . Molarity = moles of solute/Liters of solution (250 ml = 0.25 Liters ) 0.100 M C 7 H 5 N 3 O 6 = X moles/0.25 L = 0.025 moles -------------------------now, 0.025 moles C 7 H 5 N 3 O 6 (227.14 grams/1 mole C 7 H 5 N 3 O 6 ) = 5.68 grams TNT ====a good firecracker!
95% means 95ml of every 100 is alcohol. So the answer is 228ml (take away 1/20, i.e. 12ml)
How many milliliters of a 50 percent acid solution and how many milliliters of a 20 percent acid solution must be mixed to produce 36 mL of a 30 percent acid solution?
A nswer:. 12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solution. Let x = 50% acid solution. y = 20% acid solution. Equations:. x + y = 36mL ----equation (1). 0.5x + 0.2y = 0.3 * 36. 0.5x + 0.2y = 10.8. multiplying by 10. 5x + 2y = 108 ----equatio…n (2). eliminating equations (1) and (2). -2(x + y = 36)-2. -2x -2y = -72. 5x +2y = 108. =. 3x = 36. x=12. substitute x=12 to equation (1). 12 + y = 36. y = 36 - 12. y = 24. thus 12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solution. (MORE)
500ml = 500cm3 = 0.5dm3 0.250M = 0.250mol/dm3 number of moles = molarity x volume number of moles = 0.250mol/dm3 x 0.5dm3 = 0.125mol 0.125mol of NaCl is needed to prepare the required solution.
By definition, 1 liter of a 4 M solution must contain 4 moles of the solute. Because solutions are homogeneous mixtures and 250 ml is one quarter of a liter, the amount of KCl required is 1 mole. The gram molar mass* of KCl is 74.55 grams; this is therefore the amount of KCl required.** ____________…______________________ *Because KCl is an ionic rather than a molecular compound, its "gram molar mass" should preferably be called its "gram formula unit mass". **Strictly according to the rules of significant digits, this answer should be written as "7 X 10" grams, because the datum "4M" needed to calculate the answer contains only one significant digit. (MORE)
How much water must be added to 40 ml of a 25 percent by weight solution to make a 2 percent by weight solution?
About 80ml of water must be added to 40ml of a 25 percent by weightsolution to make a 2 percent by weight solution.
1 ml of water has a mass of approx 1 gram so 50 ml = approx 50 grams. Suppose x grams of sugar are required for a 3% (by mass) solution. Therefore, x/(50+x) = 3/100 That is 97x = 150 so that x = 150/97 = 1.546 grams, approx.
0.1 M Ca(OH) 2 = 0.1 mol/L = 0.01 mol/100mL = 0.01( mol /100mL) *74.093(g/ mol ) = 0.74 g/100mL Ca(OH) 2