ASCP Lab math
Determine the molarity of a solution of KOH that is contained in a Class A 1-L volumetric flask filled to the calibration mark. The content label reads 18 g KOH. The gmw of KOH=56.1 g/mol.
(18 g KOH/ L) x (1 mol/ 56.1 g HCI) = 0.32 mol/L
How many milliliters of a 5 M solution would be needed to make 250 mL of a 2 M solution?
To solve: (V1) (M1) = (V2) (M2) Where V = volume and Molarity (V1)(5M) = (250 mL) (2M) 5V1 = 500 mL V1 = 100 mL (or 100 mL 5M solution) Therefore, when 100 mL of a 5M solution is diluted to 250 mL, the resulting solution is 2M.
An aqueous 3.00 M KOH solution is prepared with a total volume of 0.650 L. The molecular weight of KOH is 56.1056 g/mol. What mass of KOH (in grams) is needed for the solution?
109.4 g Molar concentration is defined as moles (mol) of solute per liter (L) of solution. Molar concentration= mol solute / L solution Rearrange this equation to solve for the moles of solute. Input the molar concentration and volume will tell you the moles of HCI. mol solute= molar concentration XL solution =3.00 M X 0.650 L =1.95 mol The moles of HCl then need to be converted to grams using the molecular weight. g HCl = 1.95 mol X (56.1056 g/l mol) =109.49
In a moderately bloody specimen, 250 cells are counted in 1 primary square. The dilution is 1:100. What is the count/UL?
250*100/0.1 = 2.5 * 10^5
A 200 mg/dl solution was diluted 1:10. This diluted solution was then additionally diluted 1:5. What is the concentration of the final solution?
After the first dilution where 1 part of the 200 mg/dL solution was diluted with 9 parts of diluent, your left with 20 mg/dL of the original solution. Said another way, if the original solution of 200 mg/dL if diluted 1:10, then what your left with is 200 mg/dL divided by 10 which equals 20 mg/dL. After another dilution, 1 part of the 20 mg/dl solution is now diluted with 4 parts of diluent, leaving only 4 mg/dL of the original solution left. Or you can think of the new diluted solution of 20 mg/dL is diluted 1:5, therefore what you are left with is 20 mg/dL divided by 5 which equals 4 mg/dl.
A 1:100 dilution of a patient's peripheral blood was made and a total of 136 platelets were counted in 5 squares of the RBC area of a Neubauer-ruled hemocytometer. What is the Platelet count? Note: The volume correction factor is 50 µL.
Count = [(Dilution Factor) x (Total cell count in 5 RBC squares) x (Volume Correction Factor) Count = (100 x 136) x 50 = 680,000 or 680 x 10^3/ µL
What is the normality (N) of a 250 ml solution that contains 6 g of NaOH (gmw=39.997 g/mol, valence=1)?
First, you need to find the equivalent weight for NaOH. The gram molecular weight (gmw) of NaOH provided is 39.997 g/mol and a valence of 1. To find the equivalent weight, you take the gmw divided by the valence, which equals 39.997 g per equivalent weight. To find the normality of a 250mL solution that contains 6 g NaOH, you need to first identify your units. Normality is expressed as equivalents per liter (Eq/L). Next, you need to look at the units you have in this case milliliters and grams. Remember NaOH equals 39.997 g per equivalent weight. Rearrange the equation so that like terms cancel out and leave Eq/L. (6 g NaOH/250 mL) x (1 Eq/39.997 g NaOH) x (1,000 mL/1 L)=0.600 Eg/l = 0.600 N
How do you make up 275 mL of a 4.6 M solution of HCI (gmw336.46 g/mol)?
Molarity (M) is expressed in units of moles per liter (mol/L). Remember that 1 mole of substance is equal to the gram molecular weight (gmw) of that substance. For this problem, you need to first determine what units are needed. In this case, the units needed are grams (g). Next, set up the equation and cancel out like units. Then, perform the appropriate calculations: (36.46 g HCl/mol) X (4.6 mol HCI / L) X (275 mL X 1 L/ 1000 mL) = 46.1 g HCI
What is the glomerular filtration rate for a patient with a serum creatinine of 2 mg/dL, if the urine creatinine was 124 mg/dL and the urine volume was 2.2 L/24 hr?
One method of calculating a glomerular filtration rate is using creatinine and urine volume to determine creatinine clearance. The equation is as follows: Creatinine Clearance = (urine creatinine X urine flow rate) / plasma creatinine; where urine flow rate = volume in mL /24 hours x h/60 min) In this case = creatinine clearance = 124 X (2200/24 x hour/60) / 2 = 94.7 or 95 mL/min
What is the standard deviation of the following data set, rounded to the nearest whole number? 97 15, 16, 18, 16, 14, 10, 20, 12, 14
The mean for this set of numbers is 15. Calculate the deviation from the mean for each point: 0,1,3,1,1,5,5,3,1 square those results: 0,1,9,1,1,25,25,9,1 calculate the sum: 72 divide by the number of points minus one: 9 - 1= 8 72/8 = 9 Finally, determine the square root: 3 The standard deviation is therefore 3.
An Rh negative mother has just given birth to an Rh positive baby after 18 hours of strenuous labor. Her rosette test was positive. Upon performing the Kleihauer-Betke stain procedure, the percentage of fetal cells is found to be 1.9%. The mother's total blood volume is 5,000 mL. What dose of Rh lg (RhoGam) should be administered to the mother?
KB% x blood volume = volume of baby blood In this case: 1.9% x 5,000mL = 95 ml baby blood in maternal circulation 95mL / 30 ml per Rh Ig vial = 3.17 vials This equals 3 vials (after rounding), with the addition of 1 extra vial = the mother should have 4 vials of Rhlg administered.
After experiencing extreme fatigue and polyuria, a patient's metabolic panel is analyzed in the laboratory. The result of the glucose is too high for instrument to read. The laboratorian performs a dilution using 0.25 ml of patient sample to 750 ML of diluent. The result now reads 325 mg/dL. How should the techologist report this patient's glucose result?
The correct answer for this question is 1300 mg/dL. The laboratorian performed a 1:4 dilution by adding 0.25 ml (or 250 uL) of patient sample to 750 ul of diluent. This creates a total volume of 1000 pL. So, the patient sample is 250 ul of the 1000 uL mixed sample, or a ratio of 1:4. Therefore, the result given by the chemistry analyzer must be multiplied by a dilution factor of 4. 325 mg/dL x 4 = 1300 mg/dL.
Your laboratory has 17,264 hours of total paid labor time. Of this amount, 1458 hours are nonproductive hours. If your employees work 40-hour work weeks each year, what is the total number of FTES (full time equivalents) needed to run the laboratory?
8.3 When calculating FTES, you can divide the number of nonproductive and productive hours by 2080 (which is the total number of hours worked per year in a 40 hour workweek). In this case: FTES Calculation = 17,264 hours paid for labor / 2080 hours per person = 8.3 FTES
General Hospital is considering the addition of a new chemistry panel containing 12 tests. The laboratory is asked to calculate the total cost of quality control per new chemistry test panel. Quality control must be performed 3 times per day (every 8 hours). The labor cost per quality control test for this panel is $2.63. A month's worth of quality control reagent costs $354.00. What is the total quality control cost per new chemistry test panel if 76,000 of these new panels are performed each year?
$0.09 This scenario's answer can be calculated by first deciding what the total quality control labor costs are as well as what the total consumable costs are. In this case, if quality control is run 3 times per day, a total of 1095 quality control runs are performed each year. The direct labor cost of $2.63 multiplied by quality control runs equals $2879.85 per year in quality control direct labor. The hospital pays $354.00 per month on quality control consumables, which equals $4248.00 per year. The total quality control costs in a year are equal to $2879.85 + $4248.00 = $7127.85. If 76,000 new chemistry tests panels are performed each year, the total quality control cost per new chemistry test panel will be $0.09.
An aqueous 2.00 M HCl solution is prepared with a total volume of 0.475 L. The molecular weight of HCl is 36.46 g/mol. What mass of hydrochloric acid (in grams) is needed for the solution?
34.6 g mol solute= molar concentration XL solution =2.00 M X 0.475 L =0.95 mol The moles of HCl then need to be converted to grams using the molecular weight. g HCl = 0.95 mol X (36.46 g/l mol) =34.6 g
A 45-year-old male of average height and weight had a serum creatinine of 1.5 mg/dL and urine creatinine was 120 mg/dl; the total volume of urine collected over a 24-hour period was 1,800 mL. Calculate the creatinine clearance for this patient in mL/min.
The creatinine clearance for this patient is 100 mL/min. Creatinine clearance values are calculated using the following equation: Creatinine Clearance (ml / min) = (Urine Creatinine / Serum Creatinine) x Urine Volume (mL) / [ time (hr) x 60] For this patient Creatinine clearance (mL/min)= (120/1.5) x (1800 / [24 x 60]) 80 x 1800/1440, which is 80 x 1.25, or 100 mL/min
Proficiency Testing Service (e.g., CAP) reported the following results for a Cholesterol test (unknown Control) reported by your laboratory. Your results were compared with 648 other labs using the same type instrumentation and methodology: Your Result = 168 mg/dl Group Mean = 155 mg/dl Group SD = 11 mg/dl (SD = One Standard Deviation) Group + 2 SD Range = 22 mg/dl Calculate the value for your Standard Deviation Index and choose the correct answer below:
The standard deviation index is the number of group standard deviations that your mean differs from the group mean. The calculation is as follows: SDI = (Your Result - Group Mean) / Group Standard Deviation SDI = (168 mg/dL - 155 mg/dL)/11 = 1.18
24-hour urine collection requires HCl as a preservative. The procedure manual states the required concentration of HCl in percent rather than in normality. The laboratory only has 4N HCl available. What is the percent concentration? (HCl has a positive valence of 1)
To determine what percent concentration of HCl is needed to make the required concentration, you will need to first calculate the eq wt of HCl then use the following equation to determine the percent concentration of HCl: N = % x 10) / eq wt. eq wt = mol wt / pos valence eq wt = 36.5 / 1 eq wt = 36.5 N = (% x 10) / eq wt 4N = (% x 10) / 36.5 X% x 10 / 10 = 146 / 10 X% = 14.6% HCI
How would one prepare 3 ml of a 5% albumin working solution from a stock 30% albumin solution?
0.5 mL stock 30% albumin + 2.5 mL diluent Use the following formula to determine the amount of stock solution to use in preparing a working solution of lesser concentration: (% Stock solution) x (vi) = (% Working solution) x (V2) V1 = volume of the stock solution V2 = volume of the working solution Then: (30) x (V1) = (5) x (3 mL) 30 V1 = 15 mL V1 = 15 mL /30 = 0.5 mL total volume of working solution that is needed is 3 mL. Therefore, you would add 0.5 mL stock albumin to (3ml total volume working solution - 0.5 mL stock solution).
Given the following information, calculate the results in mg/24 hrs for a 24 hour urine protein. Total volume for 24 hours = 2400 mL Urine protein = 2.7 mg/dL
2.7 mg/dL X 2400 mL/24 hr X 1 dL/100 mL = 2.7 mg X 2400/100 64.8 mg/24 hr
The weight of NaCl present in 650 ml of a 0.85% NaCl solution is:
Weight (in grams) of substance in a quantity of solution = (total ml volume / 100 mL) x (percent concentration) So: Weight = (650 mL / 100 mL) x (0.85%) = 5.525 grams