What are the mole fractions of hno3 and water in a concentrated solution of nitric acid (68.0% hno3 by mass)? (a) outline the steps necessary to answer the question. (b) answer the question

mol fraction HNO₃ = 0.38

mol fraction H₂O    = 0.62

Explanation:

The mole fraction of a solution is an expression of concentration given by

Xₐ = nₐ / nt

where nₐ is the number of moles of component A, and

nt is the total number of moles ( solute + solvent) present.

The number of moles we can calculate by dividing mass into molecular weight, and since we are given the concentration in percent by mass we can use this information to calculate the number of moles of the two components in the solution.

Assume 100 g of concentrated solution, so we have 68 g of HNO₃ and ( 100 - 68 ) g of H₂O.

n HNO₃ = 68 g / 63.01 g/mol = 1.08 mol HNO₃

n H₂O =     32 g/ 18.02 g/mol = 1.78 mol H₂O

nt = 1.08 mol + 1.78 mol = 2.86 mol

Now we can calculate the mol fractions:

X HNO₃ = 1.08 mol / 2.86 mol = 0.38

X H₂O =    1.78 mol / 2.86 mol = 0.62

( since the solution is binary we could also calculate the mol fraction of H₂O as 1 - X HNO₃ )

the correct answer is boyle

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.