Lec 1: Laboratory Mathematics Flashcards
How many grams of NaCl will be needed to prepare 100mL of a 10% (w/vol.) solution?
10% = (g of NaCl/100mL) x 100 10% x 100mL = 10 grams of NaCl
How many grams of NaOH will be needed to prepare 200mL of a 15% (w/vol.) solution
15% = (g of NaOH/200mL) x 100 15% x 200mL = 30 grams of NaOH
How much HCl will be needed to prepare 1L of a 0.25 solutionM?
○ MW=36
○ SG = 1.19
○ %Assay = 37%
Also the liquid form.
0.25M = (x)/(36g x 1L)
0.25M x (36g x 1L) = 9 grams of HCl
Further conversion: 9 grams/ 1.19 = 7.56 mL 7.56 mL/37% = 20.44mL of liquid HCl
How much CH3COOH will be needed to prepare 800mL of a 0.5M solution?
○ MW=60
○ SG = 1.06
○ %Assay = 99%
+ liquid form
0.5M = (x)/(60g x 0.8L)
0.5M x (60g x 0.8L) = 24 grams of
CH3COOH
Further conversion: 24 grams/1.06 = 22.64 mL
22.64mL/99% = 22.87mL of liquid CH3COOH
How much H2SO4 will be needed to prepare 1L of a 0.35N solution?
○ MW=98
○ SG = 1.84
○ %Assay = 96%
+ liquid form
0.35N = (x)/(49g x 1L)–sulfuric acid has 2 replaceable hydrogens
0.35M x (49g x 1L) = 17.15 grams of H2SO4
Further conversion: 17.15 grams/1.84 = 9.32mL 9.32mL/96% = 9.71mL of H2SO4
How much HNO3 will be needed to prepare 500mL of a 0.5N solution?
○ MW=63
○ SG = 1.42
○ %Assay = 70%
+ liquid form
0.5N = (x)/(63g x 0.5L)
0.5N x (63g x 0.5L) = 15.75 grams of
HNO3
Further conversion: 15.75 grams/1.42 = 11.09mL
11.09mL/70% = 15.84mL of HNO3
What will be the final volume if the 100mL 10% NaCl you prepared is to be reduced to 5%?
Using the C1V1 = C2V2 formula: (10%)(100mL) = (5%)(x)
x = (10%)(100mL)/5%
x = 200mL
Convert 0.35N of H2SO4 into M
Normality = Molarity x Valence M = 0.35/2
M = 0.175
Convert the prepared 10% NaCl into M
M = (%w/v x 10)/ MW M = (10% x 10)/58.44g
M = 0.0172
Convert 325 mg/dL of Na into mEq/L
mEq/L = (mg/dL x 10)/Eq. weight mEq/L = 325 x 10/23 (atomic mass of Na) mEq/L = 141.30
Convert 100 mg/dL of glucose into mmol/L (MW = 180)
mmol/L = (100 x 10)/180
mmol/L = 5.56
Convert 5 mEq/L of Ca into mmol/L
mEq/L = mmol/L x V
mmol/L = (5 mEq/L)/2 (Ca is divalent) mmol/L = 2.5
If 0.9mL of NSS is added to 0.1ml of serum, what is the dilution?
D = 0.1mL/(0.9mL + 0.1mL) D = 0.1mL/1mL
D = 1/10 or 0.1 (look for the LCF for the numerator and denominator)
How much serum is needed to prepare a 1⁄5 dilution with a volume of 2ml?
1/5 = (x)/(2mL) x = (1⁄5) x 2mL
x = 0.4mL
A 0.5mL serum sample with a glucose value of 100mg/dL is diluted with 0.5mL NSS (Tube 1). The sample from Tube 1 is subsequently diluted with 1mL of NSS Tube 2
- What is the initial dilution?
- What is the solution dilution in Tube 2?
- What is the final concentration of
glucose?
What if there was a Tube 3 with the ff. specifications:
● Tube 1 still has 100mg/dL of glucose
● Tube 3 has taken 1mL from Tube 2 and
added 1mL of NSS
Answer to 1: D = 0.5mL/0.5mL + 0.5mL ■ D = 0.5mL/1mL
D = 1/2
Answer to 2: D = 0.5mL/1mL + 0.5mL D = 0.5mL/1.5mL
D = 1/3 x (1/2)–multiply to the previous dilution
D = 1/6
Answer to 3: 100mg/dL x 1⁄6 = 16.67mg/dL
Answer: D = 1mL/1mL + 1mL D = 1/2
D=1⁄2x1⁄3x1/2
D = 1/12 or 0.083
Final Concentration = 100mg/dL x 1/12 Final Concentration = 8.33mg/dL