MANNITOL 15% Ressources

Application Information

This drug has been submitted to the FDA under the reference 016080/003.

Names and composition

"MANNITOL 15%" is the commercial name of a drug composed of MANNITOL.

Forms

ApplId/ProductId Drug name Active ingredient Form Strenght
016080/003 MANNITOL 15% MANNITOL INJECTABLE/INJECTION 15GM per 100ML
016269/003 MANNITOL 15% MANNITOL INJECTABLE/INJECTION 15GM per 100ML
016472/005 MANNITOL 15% MANNITOL INJECTABLE/INJECTION 15GM per 100ML

Similar Active Ingredient

ApplId/ProductId Drug name Active ingredient Form Strenght
016080/006 MANNITOL 10% W/ DEXTROSE 5% IN DISTILLED WATER MANNITOL INJECTABLE/INJECTION 10GM per 100ML
020006/001 MANNITOL 5% IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 5GM per 100ML
020006/004 MANNITOL 20% IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 20GM per 100ML
016080/005 MANNITOL 15% W/ DEXTROSE 5% IN SODIUM CHLORIDE 0.45% MANNITOL INJECTABLE/INJECTION 15GM per 100ML
020006/002 MANNITOL 10% IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 10GM per 100ML
005620/001 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML
013684/003 OSMITROL 20% IN WATER MANNITOL INJECTABLE/INJECTION 20GM per 100ML
019603/004 MANNITOL 20% IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 20GM per 100ML
019603/001 MANNITOL 5% IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 5GM per 100ML
020006/003 MANNITOL 15% IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 15GM per 100ML
016080/007 MANNITOL 5% W/ DEXTROSE 5% IN SODIUM CHLORIDE 0.12% MANNITOL INJECTABLE/INJECTION 5GM per 100ML
014738/001 MANNITOL 20% MANNITOL INJECTABLE/INJECTION 20GM per 100ML
016080/001 MANNITOL 5% MANNITOL INJECTABLE/INJECTION 5GM per 100ML
016080/002 MANNITOL 10% MANNITOL INJECTABLE/INJECTION 10GM per 100ML
016080/003 MANNITOL 15% MANNITOL INJECTABLE/INJECTION 15GM per 100ML
016080/004 MANNITOL 20% MANNITOL INJECTABLE/INJECTION 20GM per 100ML
016269/001 MANNITOL 5% MANNITOL INJECTABLE/INJECTION 5GM per 100ML
016269/002 MANNITOL 10% MANNITOL INJECTABLE/INJECTION 10GM per 100ML
016269/003 MANNITOL 15% MANNITOL INJECTABLE/INJECTION 15GM per 100ML
016269/004 MANNITOL 20% MANNITOL INJECTABLE/INJECTION 20GM per 100ML
016269/005 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML
016269/006 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML
016472/002 MANNITOL 10% MANNITOL INJECTABLE/INJECTION 10GM per 100ML
016472/004 MANNITOL 20% MANNITOL INJECTABLE/INJECTION 20GM per 100ML
016472/005 MANNITOL 15% MANNITOL INJECTABLE/INJECTION 15GM per 100ML
013684/005 OSMITROL 5% IN WATER IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 5GM per 100ML
013684/006 OSMITROL 10% IN WATER IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 10GM per 100ML
013684/007 OSMITROL 20% IN WATER IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 20GM per 100ML
013684/008 OSMITROL 15% IN WATER IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 15GM per 100ML
016704/002 RESECTISOL MANNITOL SOLUTION/IRRIGATION 5GM per 100ML
022368/002 ARIDOL MANNITOL POWDER/ INHALATION 10MG
022368/003 ARIDOL MANNITOL POWDER/ INHALATION 20MG
022368/004 ARIDOL MANNITOL POWDER/ INHALATION 40MG
013684/001 OSMITROL 5% IN WATER MANNITOL INJECTABLE/INJECTION 5GM per 100ML
013684/004 OSMITROL 15% IN WATER MANNITOL INJECTABLE/INJECTION 15GM per 100ML
013684/002 OSMITROL 10% IN WATER MANNITOL INJECTABLE/INJECTION 10GM per 100ML
022368/001 ARIDOL KIT MANNITOL POWDER/INHALATION N per A,5MG,10MG,20MG,40MG
016772/002 RESECTISOL IN PLASTIC CONTAINER MANNITOL SOLUTION/IRRIGATION 5GM per 100ML
019603/003 MANNITOL 15% IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 15GM per 100ML
019603/002 MANNITOL 10% IN PLASTIC CONTAINER MANNITOL INJECTABLE/INJECTION 10GM per 100ML
080677/001 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML
083051/001 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML
086754/001 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML
087409/001 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML
087460/001 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML
089239/001 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML
089240/001 MANNITOL 25% MANNITOL INJECTABLE/INJECTION 12.5GM per 50ML

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Answered questions

A solution contains 15.0 g of mannitol, C6H14O6, dissolved in 500g of water at 40degrees C?
A solution contains 15.0 g of mannitol, C6H14O6, dissolved in 500g of water at 40degrees C. The vapor pressure of water at 40 degrees C is 55.3mmHg. A) calculate the vapor pressure of the solution (assume mannitol is a nonvolatile solute) B) calculate the boiling point of the solution Asked by Lorinda Vates 1 month ago.

moles mannitol = 15.0 g / 182.162 g/mol = 0.0823 moles water = 500 g / 18.02 = 27.7 mole fraction water = 27.7 / 27.7 + 0.0823 =0.997 molality = 0.0823 / 0.500 Kg water =0.165 A) p = p°X p = 55.3 x 0.997 = 55.1 mm Hg B) delta T = kb x m = 0.512 x 0.165 = 0.0845 boiling point = 100.0845 °C Answered by Elvin Kovacik 1 month ago.


The IV dose of mannitol is 1.5 g/kg, administered as a 15% solution. How many mls of the solution should be ?
administered to a 150 pound patient? Asked by Christia Hetcher 1 month ago.

15% = 15gm / 100ml Convert the weight to Kg 150lb * (1Kg / 2.2lb) = 68.2Kg Setup (always include units!) (68.2Kg) * (1.5gm/1Kg) * (100ml/15gm) cancel out the Kg and gm, you are left with ml, the units you want Do the math (68,2 * 1.5 * 100) / (1 * 15) 10230 / 15 Answer = 682ml Email me if you have any questions or problems EDIT: Agreed, John. From a practical view that's how I would do it, but these instructors just see it as a math problem. Answered by Luigi Ivancevic 1 month ago.

Sometimes, there's actually sense in things. In this case, the mannitol comes as 1.5 grams per 10 ml, so you only need round off the weight (70 kg is only four pounds off), multiply by ten, and give 700 ml. Answered by Tommie Pillitteri 1 month ago.

Mannitol Dose Answered by Jocelyn Lecrone 1 month ago.


The IV dose of mannitol is 1.5/kg,administred as a 15% solution.How many mls of the solution should be adminis?
administered toa 150 pound patiend? Asked by Fred Rhoad 1 month ago.

Round off your patient's weight to 70 kg and it works itself out. By the way, you left out an important part: 1.5 grams per kg would be a very high dose (700 ml), and 1.5 ml/kg (15 grams) a conservative dose. Answered by Darrel Helton 1 month ago.


The IV dose of mannitol is 1.5g/kg, administered as a 15% solution. How many mls of the solution should be adm
The IV dose of mannitol is 1.5g/kg, administered as a 15% solution. How many mls of the solution should be administered to a 150 pound patient? Asked by Shirly Ylonen 1 month ago.

It's allowable to round to 70 kgs = 154 lbs. We'l use that. 70 x 1.5g = 105 grams You want 105 grams in solution. 700 ml x 0.15g/ml = 105 grams Answered by Magaret Buzo 1 month ago.

You really need to learn how to do these simple equations yourself. If your patient weighs 150 lbs, then they weigh 68.18 kgs. If the mannitol is a 15% solution, then it contains 15 gm per 100 mL, or 0.15 g/mL. if the dose is 1.5 gm per kg then you multiply that by 68.18 to get the total dose 102.7 gm. If each mL only contain 0.15 gm then you divide 102.7 by 0.15 to get 681.8 mL. You would need to administer 681.8 mL of mannitol to give the right dose of 102.7 gm. And no, with some drugs, it's not acceptable to round up due to narrow toxicity windows and most especially if you have a pediatric patient. so although the math is easier on the 70 kg example, it's not correct for this individual patient. Answered by Faye Lauser 1 month ago.

You have some scary answers! Your dose is about 100 grams, and the solution is 15 grams per deciliter, so you'll need 700 ml. With this osmotic diuretic, it's perfectly fine to round off (even in the newborn) and 10 ml/kg is a heck of a lot easier to remember. Answered by Arnold Schnicke 1 month ago.

Well 150 lbs = 68.18kgs. So ..I guess 102.27gs? Answered by Lady Heimbach 1 month ago.


Chemistry problem help about vapor pressure?
A solution contains 15.0g of mannitol C6H1406, dissolved in 500g of water at 40C. The vapor pressure of water at 40 C is 55.3mmg A. Calculate the vapor pressure of the solution (Assume mannitol is a nonvolatile solute) B. Calculate the boiling point of the solution Asked by Leopoldo Halcon 1 month ago.

Part A requires the use of Rault's law, which states that the vapor pressure of a solution will be the vapor pressure of pure solvent times the mole fraction of solvent in the solution. P = Xsolv x P0 Xsolv is the mole fraction of the solvent - the number of moles of solvent divided by the total number of moles: solvent plus solute. Moles of H2O = 500g x (1mole/18g) = 27.78 moles Moles mannitol = 15.0 g x (1 mole/182g) = 0.0824 moles. Total moles = 27.86 mole fraction of H2O = 27.78/27.86 = 0.997 Solution vapor pressure = 0.997 x 55.3 mmHg = 55.13 mmHg b). The equation for boiling point elevation is DeltaT = m x Kb, where m is the molality of the solution (moles/kg of solvent), and Kb is the boiling point elevation constant, which 0.52 deg.C/m for water). molality = 0.0824 moles/500g = 0.1648 m Change in temperature = 0.52 degC/m x 0.1648m = 0.085 degrees C. In other words, the solution boils at 100.085 degrees. Seems like a very small change, but if you think about it, 10.5 g is not many moles of mannitol. Answered by Jose Tuerk 1 month ago.


What are the ingredients of mannitol salt agar?
what is the composition of mannitol salt agar Asked by Qiana Bicknase 1 month ago.

Typically Peptone ........................................... 10.0 g Beef Extract ....................................... 1.0 g D-Mannitol ........................................ 10.0 g Sodium Chloride ................................ 75.0 g Agar ........................................... 15.0 g Phenol Red ....................................... 25.0 mg per litre of water,which is then sterilised in an autoclave and poured in to plates before setting Answered by Jenniffer Loeber 1 month ago.

Composition Of Mannitol Salt Agar Answered by Rachel Rhynes 1 month ago.


Chemistry problem? Help PLEASE!?
nope, nothing else was provided for this problem... that's what i was thinking too! =\ Asked by Gwenn Mentzel 1 month ago.

A hospital medication order calls for the administration of 100 g of mannitol to a patient as an osmotic diuretic over a 24-hour period. If 27.8 milliliters/hour of a 15% (w/v) mannitol injection are administered to a patient, a. How many milliOsmoles of mannitol (Mw=182 g/mol) would be present in the prescribed dosage? b. What is the osmolarity of the mannitol solution? Answered by Eleanor Petersdorf 1 month ago.

Additional info: I am at work and the Mannitol bottle for a 50% solution is 1.373 mOsmol per ml. I am not sure if there is a way to directly calculate what a 15% solution would be. I would guess that the osmolality would be (15%/50% x 0.373) + 1 which would be 1.1119 per ml. 15% mannitol has 15 grams per 100 ml so 100 grams would be 667 ml. So, 1.1119 x 667 ml = 741.64 mOsmol = 100 grams of mannitol. So, this is the best guess I can give you. Let me know if it turns out to be right. Thanks, David Answered by Wynell Afurong 1 month ago.

Mannitol Mw Answered by Trang Berney 1 month ago.


If an intravenous solution contains 2 percent mannitol?
How many milliliters of the solution should be administered to provide a patient with 30 grams of mannitol? Asked by Kris Valesquez 1 month ago.

Density of water = 1 gram per cm3 2% is 2 g per 100g of water or solution (unspecified in the question). However, it doesn't matter at all for the accuracy required. Let us assume that in each 100 g of solution there are 2 g of Mannitol, and that the density of the solution is 1 g per litre. Accurate enough for this dilute solution. We need 15 times more solution therefore we need 100X15 g of water = 1,500g of water giving 1.5 litres of water and 1.5 litres of solution to the accuracy required. Answered by Nydia Kinkead 1 month ago.


Need help on these problems and a chemistry concept?
And is this true: Whenever I multiply two numbers I would cross out L And whenever I divide two numbers I would cross out mol? Asked by Inge Pflieger 1 month ago.

I can not understand these two problems at all: 1. In a 5.0% (m/v) glucose solution. How many L of a 5.0% (m/v) glucose solution would you need to obtain 55g of glucose? 2. A patient receives 100 ml of a 20% (m/v) mannitol solution every hour. How many grams of mannitol does the patient receive in 15 hours? 3. Also I solved a problem. But the units are NOT adding up. M is not the same thing as moles. I don't understand what I did exactly: What volume of 3.00 M KCL will contain 15.3 g of KCL? first I converted grams of KCL to moles. then I did 0.205 mol KCL divided by 3.00 M KCL. . I got liters and converted it to ml and got the right answer apparently. But how does mol and M cancel? I don't understand that. And when I do these problems how do I know when I cross multiply or if I divide because in that problem I divided. However in this problem I did not divide: How many grams of sodium chloride are needed to prepare 125 ml of 0.045 M NaCl solution? 0.045 M= x divided by 0.125 L. I multiplied 0.045 and 0.125 and got moles and converted it to grams and I got the right answer. But I cross multiplied, I didn't divide. Answered by Adrian Kushnir 1 month ago.

5% (m/v) means in 100 mL, 5 g is glucose. If you need 55 g of glucose, you need 55 g / 0.05 = 1100 mL or 1.1 L. You can also work it out from 100 mL contains 5 g of glucose. 55 g is 11x greater so you need 11 * 100 mL = 1100 mL = 1.1 L 2. 100 mL * 0.2 = 20 g of mannitol. Since the 100 mL is per hour, the 20 g of mannitol is per water. In 15 hours, they get 20 g mannitol/hr * 15 hours = 300 g 3. M is not moles. M is mole/L. If you have 10 moles of something disolved in 2 L, the concentration is 10 moles / 2 L = 5 mole/L = 5 M. < first I converted grams of KCL to moles > Correct. And 0.205 moles is correct. 0.205 mol KCL / 3.00 M KCL is the same as 0.205 mol KCl / 3.00 mole/L = 0.0684 L Why does mole / mole/L = L? mole / (mole/L). now, multiply top and bottom by L giving you mole * L / (mole/L * L). The moles cancel out and the two L terms in the bottom cancel out leaving you with L. < And when I do these problems how do I know when I cross multiply or if I divide because in that problem I divided > I'm not sure what you are asking. Watch your units and make sure they cancel out correctly, that will catch a lot of mistakes if you multiply when you should have divided. < How many grams of sodium chloride are needed to prepare 125 ml of 0.045 M NaCl solution? > So I have mL of NaCl solution and I have concentration in M which we know is mole/L. In this case we want to multiply because that will cancel out the volume terms, mL and L (with some conversion). 125 mL * 0.045 mole/L * 1 L / 1000 mL = 0.005625 mole of NaCl. The molar mass of NaCl is 58.44 g/mole so 0.005625 mole * 58.44 g/mole = 0.328 g NaCl. Notice how again the moles cancel out leaving you with grams, what you want when trying to find mass. If you don't include numbers, you just get 125 * 0.045 / 1000 and it's really easy to get confused 'should I multiply, should I divide?" and if you get it wrong, it's anything but obvious when you try to check your math. If you include your units, you have a built in check. An example 0.125 L / 0.2 mole/L = 0.625 L^2 / mole versus 0.125 L * 0.2 mole/L = 0.025 moles. If you are trying to find the moles of something, which set of units look correct? If you don't include units, you get 0.125 / 0.2 versus 0.125 * 0.2. Which one is right? How do you know? That's why you always include your units and make sure they cancel correctly. If they don't cancel correctly, you've likely made a mistake. Answered by Adell Vandevort 1 month ago.


Calculate the vapor pressure of the solution, Calculate the boiling point of the solution.?
A solution contains 15.0 g of mannitol, C6H14O6, dissolved in 500 g of water at 40 degreeC. The vapor pressure of water at 40 degreeC is 55.3 mmHg. (a) Calculate the vapor pressure of the solution. (Assume mannitol is a nonvolatile solute.) (b) Calculate the boiling point of the solution. Asked by Georgiana Powles 1 month ago.

a. Since you know the vapor pressure of pure water, and you can estimate the vapor pressure of pure mannitol as zero, you can use Raoult's law to get the vapor pressure of the solution: P = (Xp*)water + (Xp*)mannitol where X is the mole fraction and p* is the vapor pressure of the pure substance. b. You've got a couple of ways to go here. You can use the boiling point elevation colligative property deltaT = Kb x C where Kb is the boiling point elevation constant for water and C is the concentration of mannitol in mol/kg (molALity, not molARity, right?) Or, you can use the Clausius–Clapeyron equation, starting either with the vapor pressure you just calculated, or with the vapor pressure calculated for the solution (with Raoult's law again) at the normal boiling point of pure water (100C and 1atm, right?): P exp( Hvap / RT ) = constant Once more, slowly and simply: 1. Find the vapor pressure of the *pure* substances near the temperature of interest. Here the boiling point of your solution is gonna be near the boiling point of the solvent, yeah? 2. Use Raoult's law to calculate the vapor pressure of the solution from the vapor pressures of the pure substances and their mole fractions in solution. 3. Use Clausius-Clapeyron to calculate the boiling point of the solution, which is the temperature at which total vapor pressure of the solution is equal to atmospheric pressure. Answered by Claudette Kennemore 1 month ago.

use dp/dT = DtrsH / T DtrsV since DtrsV for a gas can be approximated as Vgas-Vliq and Vgas is so much larger than Vliq the final volume will be essentially Vgas = RT/p (assuming everything is in molar quantities) dp / dT = p DtrsH / (T^2) nR dp/p = DtrsH dT / (T^2) R (integrate this equation) ln(p2/p1) = - (DtrsH / R)*(1/T2 - 1/T1) putting the values in as above we get DtrsH. Now that we know the value of DtrsH we put it back into the above equation with one of the set of values with the pressure of 760mmHg and solve for T2. Example: ln(760/542.5) = -(DtrsH /R) * ( 1/T2 - 1/343K) rearrange and get T2. Answered by Monroe Rakowski 1 month ago.


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