The absorbacne of 0.20 corresconds top the concentrated sample.
Step one - Using calibration graph: 0.20 absorbance = 0.05M
-OK
so the concentration in the original solution is
0.05 M/100=0.0005 M (the sample was concentrated by a factor of 100)
Step three - 0.0005 mol x 58.9g/mol = 2.945 = 0.0285 g
so there is 29 mg of CO2+ per litre of water.
In order to analyze the waste water containing Co^2+ from a manufacturing process, 1.0L of water was evaporated to 10.0mL, then placed in a spectrophotometer tube. The absorbance was found to be 0.20. Using your calibration curve, calculate the number of milligrams of co^2+ in 1.0L of waste water.
Step one - Using calibration graph: 0.20 absorbance = 0.05M
Step two - 0.05M x 1.0L = 0.05mol
Step three - 0.05mol x 58.9g/mol = 2.945 = 3g
Step 4 - 3g = 0.0003mg
There are 0.0003mg Co^2+ in 1.0L of water.
Can someone check if this is the answer to the question? I'm not sure if this is how I'm supposed to do. If I'm wrong, please correct me.
Thank you very much.
2 answers
mg = (1000mg/1g) × (58.9g/1mol) × (0.05mol/1L) × 0.010L = 28mg