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β 12y agofirst calculate
vd * CD =Vc * Cc Vc=50 * 2 / 5 =20ml
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β 12y agoTo prepare 100 ml of a 5% dextrose solution from a 50% dextrose solution, you would use the formula: C1V1 = C2V2. You will need 10 ml of the 50% solution (C1) and dilute it with 90 ml of water (V1) to obtain the desired 100 ml of 5% dextrose solution.
To calculate the amount of dextrose in the IV solution, you first convert 5% w/v to mg/mL. Next, you multiply the concentration by the volume of the IV solution to find the total amount of dextrose in it. For a 475 mL IV solution with 5% w/v dextrose, you would have 475 mL * 5 mg/mL = 2375 mg of dextrose.
To find the number of grams of dextrose in a 1L solution with 5.5% concentration, you can use the formula: mass = volume x concentration. So, for this case, it would be 1L x 5.5% = 55g of dextrose in a 1L solution.
Normal Saline Solution in 5% Dextrose or D5NSS is a hypertonic solution. It can be used for the temporary treatment of shock if plasma expanders are not available. However, do not administer this IV for clients with cardiac or renal conditions.
To prepare 1 liter of a 5% alcohol-water solution, you would mix 50 mL of alcohol with 950 mL of water. This ratio would give you a total volume of 1 liter with 5% of it being alcohol.
To prepare a 1% solution of table sugar, you would dissolve 1 gram of table sugar in 100 mL (or 100 grams) of water. This would give you a solution where 1% of the total weight is sugar.
In a 70% dextrose solution, 70% of the total weight is dextrose. To calculate the grams of dextrose in 400ml of this solution, you would multiply 400ml by 70% (or 0.70) to find the amount of dextrose present.
pharmacist
To calculate the amount of dextrose in the IV solution, you first convert 5% w/v to mg/mL. Next, you multiply the concentration by the volume of the IV solution to find the total amount of dextrose in it. For a 475 mL IV solution with 5% w/v dextrose, you would have 475 mL * 5 mg/mL = 2375 mg of dextrose.
To find the number of grams of dextrose in a 1L solution with 5.5% concentration, you can use the formula: mass = volume x concentration. So, for this case, it would be 1L x 5.5% = 55g of dextrose in a 1L solution.
See the two Related Questions to the left for the answer.The first is how to prepare a solution starting with a solid substance (and dissolving it). The second question is how to prepare a solution by diluting another solution.
To prepare a 6N ammonium hydroxide solution from a 30% solution, you would first calculate the molarity of the 30% solution. Then, you would dilute the 30% solution with water to reach the desired concentration. Calculate the volume of the 30% solution needed and dilute it with water to reach the final volume for a 6N concentration.
Dextrose is a solid so it doesn't really have a pH until it is in solution. Once it is in solution, it would depend on what the solvent was and the concentration of the dextrose. So it could range greatly, but most are kept around 6.4.
To prepare a 40.0% lithium nitrate solution, 60mL of the solution would be solvent (water) and 90mL would be solute (lithium nitrate). Therefore, you would need 90mL of lithium nitrate to prepare 150mL of the 40.0% solution.
To prepare a 30% NaCl solution, you would dissolve 30 grams of table salt (NaCl) in enough water to make a total solution volume of 100 mL. This would result in a 30% concentration of NaCl in the final solution.
Normal Saline Solution in 5% Dextrose or D5NSS is a hypertonic solution. It can be used for the temporary treatment of shock if plasma expanders are not available. However, do not administer this IV for clients with cardiac or renal conditions.
To prepare a 1% solution of table sugar, you would dissolve 1 gram of table sugar in 100 mL (or 100 grams) of water. This would give you a solution where 1% of the total weight is sugar.
Unfortunately sugar is an imprecise term. It is better to specify glucose (usually means dextrose) or sucrose or fructose etc. A 5 percent solution of one of these sugars would contain 5 grams weight dissolved in 100mL of water (or could be another solvent).