Answer:
Vapor pressure of solution is 78.2 Torr
Explanation:
This is solved by vapor pressure lowering:
ΔP = P° . Xm . i
Vapor pressure of pure solvent (P°) - vapor pressure of solution = P° . Xm . i
NaCl → Na⁺ + Cl⁻ i = 2
Let's determine the Xm (mole fraction) These are the moles of solute / total moles.
Total moles = moles of solvent + moles of solute
Total moles = 0.897 mol + 0.182 mol → 1.079 mol
0.182 / 1.079 = 0.168
Now we replace on the main formula:
118.1° Torr - P' = 118.1° Torr . 0.168 . 2
P' = - (118.1° Torr . 0.168 . 2 - 118.1 Torr)
P' = 78.2 Torr
Answer is: <span>the percent ionization is 0,19%.
</span>Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
Ka(HA) = 3,6·10⁻⁷.
c(HA) = 0,1 M.
[H⁺] = [A⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [A⁻] / [HA].
0,00000036 = x² / 0,1 M - x.
Solve quadratic equation: x = 0,00019 M.
α = 0,00019 M ÷ 0,1 M · 100% = 0,19%.
Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point
elevation is directly proportional to the molality of the solution
according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point
elevation.
Kb - the ebullioscopic
constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
Answer:
I don't fully understand what this is about...
Explanation:
sorry :(