Explanation:
Reaction for oxidation of gaseous nitric oxide to aqueous nitrous acid is as follows.
By adding water, we will balance the oxygen atoms on both the sides as follows.
Now, we will balance the hydrogen atoms by adding 
ions as follows.
Now, we will add 
ions on both the sides in order to neutralize ions.
..... (1)
We will combine the and ions in order to form water in equation (1) and cancelling the common terms as follows.
Now, we will balance the charge on both the sides as follows.
Thus, we can conclude that the balanced half-reaction equation is as follows.
Answer:
The molecular weight of the unknown compound is 267.7 g/mol
Explanation:
Lowering vapor pressure → Colligative property where the vapor pressure of solution is lower than vapor pressure of pure solvent
ΔP = P° . Xm
0.526 atm - 0.501 atm = 0.025atm
0.025 atm = 0.526 atm . Xm
Xm = 0.025 atm / 0.526 atm → 0.0475 (mole fraction)
Mole fraction = Moles of solute / Total moles (solute + solvent)
0.0475 = Moles of solute / Moles of solute + Moles of solvent
We determine the moles of solvent → 116.2 g . 1mol / 58 g = 2 moles
0.0475 = Moles of solute / Moles of solute + 2
0.0475 moles of solute + 0.095 = Moles of solute
0.095 = Moles of solute - 0.0475 moles of solute
0.095 / 0.9525 = Moles of solute → 0.0997 moles
Molas mass = g/mol → 26.7 g / 0.0997 mol = 267.7 g/mol
