The density of sea water is 1.03 g/ml. A sample of sea water contains 60 ppm Ca. What is the mass of Ca(NO3)2
present in 20 L of this sample, assuming that it is the only source of Ca?
To find the mass of Ca(NO3)2 present in the sample, we need to first find the mass of the sea water in the 20 L sample.
Given: Density of sea water = 1.03 g/ml
Volume of sample = 20 L
1 L = 1000 ml
So, the volume of the sample in ml = 20 L * 1000 ml/L = 20000 ml
Mass of sea water = volume * density = 20000 ml * 1.03 g/ml = 20600 g
Next, we need to find the mass of Ca(NO3)2.
Given: Ca concentration = 60 ppm
We also know that 1 ppm = 1 mg/L
So, the mass of Ca present in the sample = concentration * volume = 60 ppm * 20 L = 1200 mg
Since 1 g = 1000 mg, the mass of Ca = 1200 mg / 1000 = 1.2 g
Since Ca(NO3)2 contains 2 parts of Ca in 1 part of the compound, the mass of Ca(NO3)2 = mass of Ca / 2 = 1.2 g / 2 = 0.6 g
Therefore, the mass of Ca(NO3)2 present in 20 L of the sample is 0.6 g.