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 3 months 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 3 months 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 3 months ago.

Mannitol Dose Answered by Jocelyn Lecrone 3 months 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 3 months 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 3 months 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 3 months 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 3 months ago.

Well 150 lbs = 68.18kgs. So ..I guess 102.27gs? Answered by Lady Heimbach 3 months 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 3 months 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 3 months ago.

Composition Of Mannitol Salt Agar Answered by Rachel Rhynes 3 months 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 3 months 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 3 months ago.

Mannitol Mw Answered by Trang Berney 3 months 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 3 months 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 3 months 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 3 months 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 3 months 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 3 months ago.

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 |

ApplId/ProductId | Drug name | Active ingredient | Form | Strenght |
---|---|---|---|---|

005620/001 | MANNITOL 25% | MANNITOL | INJECTABLE/INJECTION | 12.5GM per 50ML |

013684/001 | OSMITROL 5% IN WATER | MANNITOL | INJECTABLE/INJECTION | 5GM per 100ML |

013684/002 | OSMITROL 10% IN WATER | MANNITOL | INJECTABLE/INJECTION | 10GM per 100ML |

013684/003 | OSMITROL 20% IN WATER | MANNITOL | INJECTABLE/INJECTION | 20GM per 100ML |

013684/004 | OSMITROL 15% IN WATER | 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 |

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 |

016080/005 | MANNITOL 15% W/ DEXTROSE 5% IN SODIUM CHLORIDE 0.45% | MANNITOL | INJECTABLE/INJECTION | 15GM per 100ML |

016080/006 | MANNITOL 10% W/ DEXTROSE 5% IN DISTILLED WATER | MANNITOL | INJECTABLE/INJECTION | 10GM per 100ML |

016080/007 | MANNITOL 5% W/ DEXTROSE 5% IN SODIUM CHLORIDE 0.12% | MANNITOL | INJECTABLE/INJECTION | 5GM 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 |

016704/002 | RESECTISOL | MANNITOL | SOLUTION/IRRIGATION | 5GM per 100ML |

016772/002 | RESECTISOL IN PLASTIC CONTAINER | MANNITOL | SOLUTION/IRRIGATION | 5GM per 100ML |

019603/001 | MANNITOL 5% IN PLASTIC CONTAINER | MANNITOL | INJECTABLE/INJECTION | 5GM per 100ML |

019603/002 | MANNITOL 10% IN PLASTIC CONTAINER | MANNITOL | INJECTABLE/INJECTION | 10GM per 100ML |

019603/003 | MANNITOL 15% IN PLASTIC CONTAINER | MANNITOL | INJECTABLE/INJECTION | 15GM per 100ML |

019603/004 | MANNITOL 20% IN PLASTIC CONTAINER | MANNITOL | INJECTABLE/INJECTION | 20GM per 100ML |

020006/001 | MANNITOL 5% IN PLASTIC CONTAINER | MANNITOL | INJECTABLE/INJECTION | 5GM per 100ML |

020006/002 | MANNITOL 10% IN PLASTIC CONTAINER | MANNITOL | INJECTABLE/INJECTION | 10GM per 100ML |

020006/003 | MANNITOL 15% IN PLASTIC CONTAINER | MANNITOL | INJECTABLE/INJECTION | 15GM per 100ML |

020006/004 | MANNITOL 20% IN PLASTIC CONTAINER | MANNITOL | INJECTABLE/INJECTION | 20GM per 100ML |

022368/001 | ARIDOL KIT | MANNITOL | POWDER/INHALATION | N per A,5MG,10MG,20MG,40MG |

022368/002 | ARIDOL | MANNITOL | POWDER/ INHALATION | 10MG |

022368/003 | ARIDOL | MANNITOL | POWDER/ INHALATION | 20MG |

022368/004 | ARIDOL | MANNITOL | POWDER/ INHALATION | 40MG |

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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