Question

6.3g of concentrated nitric acid is diluted by adding 68g of pure water.how many oxygen molecules are present in the solution?

Answer 1

Hey there,

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 HNO3:
This stuff is 15.8M and has a density of 1.42g/mL (thanks, google)
6.3gHNO3×1mL1.42g = 4.44 mL of solution
0.0044L×(15 molL)=∗0.0702mo≤sHNO_3∗0.0702mo≤sHNO_3 xx 63g/"mol"=4.42gHNO_3Massofwater∈solution:6.3gNitricAcidSolution−4.42gHNO_3=1.88gH_2OMo≤sH_2O∈NitricAcidSolution:1.88gxx("1mol"/"18g")=∗0.104mo≤sH_2O#**

68g of pure water
68g×(1mol18g) = 3.78 moles H2O

Total Moles:
HNO3: 0.0702 moles HNO3
H2O: 0.104 moles + 3.78 moles= 3.88 moles H2O

Total Moles of Oxygen atoms:
Each HNO3 gives 3 Oxygen atoms
0.0702 moles HNO3×3 = 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:
4.091mol×(6.02×1023atoms1 mol) = 2.46×1024 Oxygen atoms.

HOPE YOU GOT IT
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