Question

6.3 gm of conc nitric acid is diluted by adding 68 gm of pure water. How many oxygen molecules are present in the solution?

Answer 1

See Below

Explanation:

I'm going to assume you mean oxygen atoms...like singular O.
Three places you get your O atoms. HNO3, water in the HNO3 solution (it is concentrated HNO3, so it is 15.8M...which means it has water, too), and the Water you add after.

Going to just add up all the moles of oxygen-containing things:
Concentrated `HNO_3`:
This stuff is 15.8M and has a density of 1.42g/mL (thanks, google)
`6.3g HNO_3 xx "1mL"/"1.42g"` = 4.44 mL of solution
`0.0044Lxx("15 mol"/L) = **0.0702 moles`HNO_3#** 0.0702 moles #HNO_3 xx 63g/"mol"`= 4.42 g`HNO_3# Mass of water in solution: 6.3g Nitric Acid Solution - 4.42g #HNO_3`= 1.88g`H_2O# Moles #H_2O# in Nitric Acid Solution: #1.88gxx("1mol"/"18g")`= **0.104 moles`H_2O#**

68g of pure water
`68gxx("1mol"/"18g")` = 3.78 moles `H_2O`

Total Moles:
`HNO_3`: 0.0702 moles `HNO_3`
`H_2O`: 0.104 moles + 3.78 moles= 3.88 moles `H_2O`

Total Moles of Oxygen atoms:
Each `HNO_3` gives 3 Oxygen atoms
0.0702 moles `HNO_3 xx 3` = 0.211 moles O atoms
Water - each water gives 1 oxygen atom:
3.88 moles O atoms
Total: 4.091 moles O atoms

Number of O atoms:
`4.091 mol xx ((6.02xx10^23 "atoms")/"1 mol")` = `2.46xx10^"24"` Oxygen atoms