When P1/P2 = C1/C2
and C is the molarity which = moles/volume
so, P1/P2 = [(mass1/mw)/volume] / [(mass2/mw)/volume]
P1/P2 = (mass1/mw)/1.5L / (mass2/mw)/1.5L
so, Mw and 1.5 L will cancel out:
∴P1/P2 = mass1 / mass2
∴ mass 2 = mass1*(P2 / P1)
= 0.278g * (78 bar / 62 bar)
= 0.35 g
∴ the quantity of argon that will dissolve at 78 bar = 0.35 g
potassium reacts the most vigorously.
We will use the formula for freezing point depression :
but first, we need to get the molality m of the solution:
- molality m = moles of C2H5OH / mass of water Kg
when moles of C2H5OH = mass of C2H5OH/ molar mass of C2H5OH
= 11.85 g / 46 g/mol
= 0.258 moles
and when we have the mass of water Kg = 0.368 Kg
so, by substitution on the molality formula:
∴ molality m = 0.258 moles / 0.368 Kg
= 0.7 mol/Kg
and when C2H5OH is a weak acid so, there is no dissociation ∴ i = 1
and when Kf is given = 1.86 C/m
so by substitution on ΔTf formula:
when ΔTf = i Kf m
∴ ΔTf = 1 * 1.86C/m * 0.7mol/Kg
= 1.302 °C
