Answer:
0.605 molal
Explanation:
Molality indicates the solute amount in a specific solvent mass.
Let’s find the amount of benzene solute.
Mass of benzene = 13.3g
Molar mass of C6H6 = 12*6 +1*6 =72+7=78g/mol
Amount of benzene = mass/molar mass
=13.3/78
=0.1705mol
Molality = amount of solute/mass of solvent in kg
Mass of solvent = 282g = 0.282kg
Molality = 0.1705/0.282
=0.605 molal
Choice (3) is correct: they exhibit the same chemical properties. Both samples are under identical conditions, so their densities remain unchanged. Gram-formula mass is an intrinsic property of the substance and does not vary. The volume would change if the mass changes.
Answer:
0.1714 (w/w) %
Explanation:
Utilizando la ecuación:
16H+(aq) + 2Cr2O72−(aq) + C2H5OH(aq) → 4Cr3+(aq) + 2CO2(g) + 11H2O(l)
Se emplean 2 moles de ion dicromato (Cr₂O₇²⁻) para titular 1 mol de alcohol (C₂H₅OH)
35.46mL = 0.03546L de una solución de Cr₂O₇²⁻ a 0.05961M utilizada para alcanzar el punto de equivalencia en la titulación contiene:
0.03546L ₓ (0.05961 moles Cr₂O₇²⁻ / L) = 2.114x10⁻³ moles Cr₂O₇²⁻
Dado que 2 moles de dicromato reaccionan por cada mol de alcohol, los moles de alcohol en la muestra de plasma son:
2.114x10⁻³ moles Cr₂O₇²⁻ ₓ ( 1 mol C₂H₅OH / 2 moles Cr₂O₇²⁻) = 1.0569x10⁻³ moles de C₂H₅OH
Como la masa molar del alcohol es 46.07g/mol, la masa de alcohol es:
1.0569x10⁻³ moles de C₂H₅OH ₓ (46.07g / mol) = 0.04869g de C₂H₅OH
Por lo tanto, el porcentaje en masa de alcohol en sangre utilizando los 28.40g de plasma es:
(0.04869g de C₂H₅OH / 28.40g) × 100 = 0.1714 (w/w) %
