Answer is: D. Na2SO4.
b(solution) = 0.500 mol ÷ 2.0 L.
b(solution) = 0.250 mol/L.
b(solution) = 0.250 m; molality of the solutions.
ΔT = Kf · b(solution) · i.
Kf - the freezing point depression constant.
i - Van 't Hoff factor.
Dissociation of sodium sulfate in water: Na₂SO₄(aq) → 2Na⁺(aq) + SO₄²⁻(aq).
Sodium sulfate dissociates on sodium cations and sulfate anion, sodium sulfate has approximately i = 3.
Sodium chloride (NaCl) and potassium iodide (KI) have Van 't Hoff factor approximately i = 2.
Carbon dioxide (CO₂) has covalent bonds (i = 1, do not dissociate on ions).
Because molality and the freezing point depression constant are constant, greatest freezing point lowering is solution with highest Van 't Hoff factor.
Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
I need a little more context but I believe you are correct
CnH2n since the equivalent unsaturation is equal with 1
