I am culturing a N2 fixing cyanobacterium and have the following media recipe fo

Question

I am culturing a N2 fixing cyanobacterium and have the following media recipe for growing them: To make 1 L of salt water add the following to fresh water:

2.94 g NaF

17.4 g SrCl2.6H2O

6.8 g KH2PO4

0.11 g FeCl3.6H2O

0.012 g ZnSO4.7H2O

0.04 g MnCl2.4H2O

0.027 g Na2MoO4.2H2O

0.06 g CoCl2.6H2O

0.025 g CuSO4.5H2O

Add 1 mL each of the first 5 stock solutions to the seawater mixture above. Add 100 mL of the Mn, Zn, and Mo stock solutions to the seawater mixture above. Add 10 mL of the Co and Cu stock solutions to the seawater mixture above. What are the final concentrations of each trace element in the seawater solution (F, Sr, P, Fe, Zn, Mn, Mo, Co, and Cu (in mmol/L)? What is the amount of each of these constituents in g/kg?

Explanation / Answer

Prepare 1 L of stock solutions of each salt

density fo water = 1 g/ml

1 L = 1 kg

moles = grams/molar mass

molarity = moles/L of solution

then mix required volumes to make sea water sample

molarity [F] in stock = 2.94 g/42 g/mol x 1 L = 0.07 M

final concentration [F] in sea water sample = 0.07 M x 1 ml/1 L = 0.07 mmol/L

In (g/kg) [F] = 0.07 x 19/1000 = 0.00133 g/kg

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molarity [Sr] in stock = 17.4 g/158.5 g/mol x 1 L = 0.11 M

final concentration [Sr] in sea water sample = 0.11 M x 1 ml/1 L = 0.11 mmol/L

In (g/kg) [Sr] = 0.11 x 87.6/1000 = 0.0097 g/kg

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molarity [P] in stock = 6.8 g/136.1 g/mol x 1 L = 0.05 M

final concentration [P] in sea water sample = 0.05 M x 1 ml/1 L = 0.05 mmol/L

In (g/kg) [P] = 0.05 x 31/1000 = 0.00155 g/kg

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molarity [Fe] in stock = 0.11 g/270.3 g/mol x 1 L = 0.0004 M

final concentration [Fe] in sea water sample = 0.0004 M x 1 ml/1 L = 0.0004 mmol/L

In (g/kg) [Fe] = 0.0004 x 55.8/1000 = 0.023 g/kg

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molarity [Zn] in stock = 0.012 g/287.6 g/mol x 1 L = 0.000042 M

final concentration [Zn] in sea water sample = 0.000042 M x 1 ml/1 L = 0.000042 mmol/L

In (g/kg) [Zn] = 0.000042 x 65.4/1000 = 0.003 g/kg

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molarity [Mn] in stock = 0.04 g/125.8 g/mol x 1 L = 0.00032 M

final concentration [Mn] in sea water sample = 0.00032 M x 100 ml/1 L = 0.032 mmol/L

In (g/kg) [Mn] = 0.032 x 55/1000 = 0.00175 g/kg

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molarity [Mo] in stock = 0.027 g/206 g/mol x 1 L = 0.00013 M

final concentration [Mo] in sea water sample = 0.00013 M x 100 ml/1 L = 0.013 mmol/L

In (g/kg) [Mo] = 0.013 x 95.94/1000 = 0.00126 g/kg

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molarity [Co] in stock = 0.06 g/130 g/mol x 1 L = 0.00046 M

final concentration [Co] in sea water sample = 0.00046 M x 10 ml/1 L = 0.0046 mmol/L

In (g/kg) [Co] = 0.0046 x 59/1000 = 0.00027 g/kg

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molarity [Cu] in stock = 0.025 g/159.61 g/mol x 1 L = 0.00016 M

final concentration [Cu] in sea water sample = 0.00016 M x 10 ml/1 L = 0.0016 mmol/L

In (g/kg) [Cu] = 0.0016 x 63.55/1000 = 0.0001 g/kg

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