balance the chemical reaction by the oxidation reduction method Phosphorus + Sodium Hydroxide + water
Answer:
a. Xm = 0.0229
b. 0.0234 moles
c. 354.1 g/mol
Explanation:
ΔP = P° . Xm
ΔP = P° - P', where P° is vapor pressure of pure solvent and P', vapor pressure of solution-
This is the formula for lowering vapor pressure.
If we apply the data given: 523 Torr - 511 Torr = 523 . Xm
Xm = ( 523 Torr - 511 Torr) / 523 Torr → 0.0229
Xm = Mole fraction of solute → Moles of solute / Total moles (sv + solute)
We can make this equation to determine moles of solute
0.0229 = Moles of solute / Moles of solute + 1
0.0229 (Moles of solute + 1) = Moles of solute
0.0229 = Moles of solute - 0.0229 moles of solute
0.0229 = 0.9771 moles of solute → 0.0229 / 0.9971 = 0.0234 moles
Molecular mass of solute → g/mol → 8.3 g / 0.0234 mol = 354.1 g/mol
Explanation:
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i hope this is helps
<span>N2 + 3H2 → 2 </span>NH3<span> from bal. rxn., 2 moles of </span>NH3<span> are formed per 3 moles of </span>H2, 2:3 moleH2<span>: 3.64 </span>g<span>/ 2 </span>g<span>/mole </span>H2<span>= 1.82 1.82 moles </span>H2<span> x 2/3 x 17
</span>
, we need to supply it
to make it
.</em>
