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.
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?
1 answer