unit convertion ( Chemistry)

(a) 8.4 g/L NaCl into mM

Molar mass of NaCl = 23 + 35.5 = 58.5 g mol-1

Consider 1 L of NaCl solution:
No. of moles of NaCl = mass/(molar mass) = 8.4/58.5 = 0.144 mol
Molarity of NaCl = mol/V = 0.144/1 = 0.144 M = 144 mM

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(b) 3.16 x 10-6 M Fe2+ into ppm

1 ppm = 1 mg of solute / 1 L of solution
Reference: http://www.ilpi.com/msds/ref/concentration.html

Molar mass of Fe2+ = 55.9 g mol-1

Consider 1 L of the solution:
No. of moles of Fe2+ = MV = (3.16 x 10-6) x 1 = 3.16 x 10-6 mol
Mass of Fe2+ = mol x (molar mass) = (3.16 x 10-6) x 55.9 g = 17.7 mg
Concentration = 17.7 mg/L = 17.7 ppm

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(c) 6.30 mM glucose (C6H12O6) into mg/dL

Molar mass of C6H12O6 = 12x6 + 1x12 + 16x6 = 180 g mol-1

Consider 1 dL (0.1 L) of the solution:
No. of moles of C6H12O6 = MV = (6.3 x 10-3) x 0.1 = 6.3 x 10-4 mol
Mass of C6H12O6 = mol x (molar mass) = (6.3 x 10-4) x 180 g = 113 mg
Hence, concentration = 113 mg/dL

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(d) 5.0 ppm Ca2+ into μM

Molar mass of Ca2+ = 40.1 g mol-1

Consider 1 L of solution.
(1 ppm = 1 mg/L)
Mass of Ca2+ = 5.0 mg = 5 x 10-3 g = 0.005 g
No. of moles of Ca2+ = mass/(molar mass) = 0.005/40.1 = 1.25 x 10-4 mol
Concentration = mol/V = (1.25 x 10-4)/1 = 125 x 10-6 M = 125 μM

整體來說，002 的答案較好，而001 的答案錯誤較多。

(a) 大家都答對了。

2008-05-30 17:41:17 補充：
(b) 大家都答錯了。
001 計算得 176.6 g/L，誤以 ppm = 1 x 10^-6 g/L，但其實 ppm = 1 x 10^-6 g/mL，因為1 mL solution 的 mass 約是 1 g。
002 說 1 ppm = 1 mg/L 其實是對的，因為 1 mg/L = 1 x 10^-6 g/mL。但002 在由 g 轉 mg 時錯誤，Mass of Fe^2+ 應等於 0.177 g = 0.177 mg (而不是17.7 mg)。
So, 正確答案是 0.177 ppm 或 0.1766 ppm。

2008-05-30 17:41:30 補充：
(c) 001 錯，002 對。
001 計到 1134 mg/L 是對的。但 1dL = 0.1L，1 dL 的 volume 只有 1L 的 1/10，所以每 dL 的 solute 含量應 x 0.1，而不是 / 0.1。

(d) 001 錯，002 對。
001 第一錯在以為 5 ppm 等於 5 x 10^-6 g/L，其實應是 5 x 10^-6 g/mL，理由如上述。第二錯在第三、四行，由 g 轉 mol，用 mol = mass/molar mass 計，應該是 mass/40，而不是 mass x 40。

a) 8.4 g/L NaCl into mM (molecular mass of Na = 23.0, Cl = 35.5)
= 8.4g / (23.0 +35.5) mol/L NaCl = 0.144 M NaCl
= 144 mM NaCl
(b) 3.16 x 10^-6 M Fe 2+ into ppm (molecular mass of Fe2+ = 55.9)
3.16 x 10^-6 mol / L Fe2+
= 3.16 x 55.9 x10^-6 g / L Fe2+
=176.6 ppm Fe2+
(c) 6.30 mM glucose (C 6 H 12 O 6 ) into mg/dL
6.30 mM glucose
= 6.30 x (12.0 x 6+1.0 x 12 + 16.0 x 6) mg/L
= 1134 mg/L = (1134 / 0.1) mg/dL
= 11340 mg/dL
(d) 5.0 ppm Ca 2+ into μM (molecular mass of calcium = 40.0)
5.0 ppm Ca2+
= 5.0 x10^-6 g/L Ca2+
= 5.0 x 40 μmol / L Ca2+
= 200 x μM Ca2+