Question

1. A sample of water was found to contain dissolved
oxygen (0,) to the extent of 4 ppm and hardness due to
Ca2+ ions as 3 ppm. Calculate the amount of 0, and
number of Ca2+ ions in 1 litre of water (Assume density
1 g/ml).​

Answer 1

Answer:

The oxygen gas in present in 1 L of water is 0.004 grams.

The calcium ion in present in 1 L of water is 0.003 grams.

Explanation:

$ppm=\frac{\text{mass of solute}}{\text{mass of solvent}}\times 10^6$

The ppm concentration of oxygen gas in water = 4 ppm

Mass of water as solvent = m

Volume of water = 1 L =1000 mL

Density of water = 1 g/mL

m = 1g/mL × 1000 mL = 1000 g

$4ppm=\frac{\text{Mass of oxygen}}{1000}\times 10^6$

Mass of oxygen gas =

$\frac{4\times 1000 g}{10^6}=0.004 g$

The ppm concentration of calcium ions in water = 3 ppm

$3ppm=\frac{\text{Mass of}Ca^{2+}}{1000}\times 10^6$

Mass of $Ca^{2+}$ ions:

= $\frac{3\times 1000 g}{10^6}=0.003 g$