Answer:
In full volume it contain 0.12 moles.
Explanation:
Given data:
Total volume= Vt = 2.9 L
Decreased volume= Vd = 1.2 L
Number of moles of air present in decreased volume= n = 0.049 mol
Number of moles of air in total volume= n = ?
Solution:
Formula:
Vt/ Vd = n (in total volume) /n ( decreased volume)
2.9 L / 1.2 L = X / 0.049 mol
2.42 = X / 0.049 mol
X = 2.42 × 0.049
X = 0.12 mol
Answer:
Phosphoric Acid would be your answer.
Answer is: V<span>an't Hoff factor (i) for this solution is 2,26.
</span>Change in freezing point
from pure solvent to solution: ΔT =i · Kf · m.
<span>Kf - molal freezing-point depression constant for water is 1,86°C/m.
</span>m - molality, moles of solute per kilogram of solvent.
n(K₂SO₄) = 16,8 g ÷ 174,25 g/mol
n(K₂SO₄) = 0,096 mol.
m(K₂SO₄) = 0,096 mol/kg.
ΔT = 0,405°C.
i = 0,405 ÷ (1,86 · 0,096)
i = 2,26.
