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 )

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.