1) If the reference range for serum Ca2+ is 8.5—10.5 mg/dL, would a result of 2.3 mmol/L be within that range? How about a result of 4.8 mEq/L? 1 Eq = 1,000 mEq

2. One of the assays your laboratory routinely carries out requires 1.0 N H2504. You have in stock a bottle of concentrated sulfuric acid with a specific gravity of 1.84 and a purity of 97%. To achieve the target concentration of 1.0 N, how many milliliters of this acid must be diluted to a final volume of 100 mL?

3. Your laboratory uses an analytical instrument for lead (Pb) that requires calibration with four standard solutions of lead to be dissolved in 5% HNOB. To prepare one standard solution of 50.0 ug/ml, you mixed 5.0 mL of a sample with 95.0 mL of diluent (solvent). What was the concentration of manufacturer's solution in mg/dL before dilution?

1. 

2.3Lmmol?(1mmol40.08mgCa2+?)(10dL1L?)=9.2mg/dL

Yes, 2.3 mmol/L is within the reference range for serum Ca²⁺.

4.8LmEq?(2mEq1mmol?)(1mmol40.08mgCa2+?)(10dL1L?)=9.6mg/dL

Yes, 4.8 mEq/L is within the reference range for serum Ca²⁺.

2.  2.75 mL

3. 100. mg/dL

Step-by-step explanation

1. Convert 2.3 mmol/L to mg/dL to see if the former value is within the reference range.

Steps:

• Convert 2.3 mmol/L to mg/L using the molar mass of Ca (40.08 g/mol or 40.08 mg/mmol) as conversion factor.
• Convert mg/L to mg/dL using 1 L = 10 dL as conversion factor.

2.3Lmmol?(1mmol40.08mgCa2+?)(10dL1L?)=9.2mg/dL

Yes, 2.3 mmol/L is within the reference range for serum Ca²⁺.

Convert 4.8 mEq/L to mg/dL to see if the former value is within the reference range.

Steps:

• Convert mEq/L to mmol/L using 2 mEq = 1 mmol as conversion factor. Ca²⁺ has a charge of +2 so 2 equiv = 1 mol.
• Convert mmol/L to mg/L using the molar mass of Ca (40.08 g/mol or 40.08 mg/mmol) as conversion factor.
• Convert mg/L to mg/dL using 1 L = 10 dL as conversion factor.

4.8LmEq?(2mEq1mmol?)(1mmol40.08mgCa2+?)(10dL1L?)=9.6mg/dL

Yes, 4.8 mEq/L is within the reference range for serum Ca²⁺.

2. Specific gravity is the ratio of the density of a material to the density of water. Because the density of water is 1 g/mL, the specific gravity of H₂SO₄ is equal to its density, which is 1.84 g/mL. A bottle of concentrated sulfuric acid that is 97% pure means that there are 97 g H₂SO₄ in 100 g solution. Convert 100 g solution to L solution using the density of the sulfuric acid solution.

100gH2?SO4?(1.84g1mL?)(1000mL1L?)=0.0543L

The molarity of this concentrated sulfuric acid solution would be

M=VMm??=0.0543L98.04g/mol97g??=18.2M

Convert 18.2 mol/L to eq/L (or N) using 2 equiv = 1 mol. H₂SO₄ has two acidic protons, thus 1 mol H₂SO₄ = 2 equiv H⁺

18.2Lmol?(1mol2equiv?)=36.4N

Use the dilution formula to determine the volume of the concentrated H₂SO₄ that is needed to prepare 1.0 N H₂SO₄ that has a final volume of 100 mL.

M_diluteV_dilute = M_concV_conc

Solve for V_conc

V_conc = M_diluteV_dilute/M_conc = (1 N)(100 mL)/36.4 N = 2.75 mL

3. Use the dilution formula to determine the concentration of the manufacturer's solution.

M_diluteV_dilute = M_concV_conc

Solve for M_conc

M_conc = M_diluteV_dilute/V_conc = (50.0 μg/mL)(100 mL)/5.0 mL = 1000 μg/mL

Convert 1000 μg/mL to mg/dL

1000mLμg?(1000μg1mg?)(1dL100mL?)=100.mg/dL