Answer:
[H₂SO₄] = 6.07 M
Explanation:
Analyse the data given
8.01 m → 8.01 moles of solute in 1kg of solvent.
1.354 g/mL → Solution density
We convert the moles of solute to mass → 8.01 mol . 98g /1mol = 785.4 g
Mass of solvent = 1kg = 1000 g
Mass of solution = 1000g + 785.4 g = 1785.4 g
We apply density to determine the volume of solution
Density = Mass / volume → Volume = mass / density
1785.4 g / 1.354 g/mL = 1318.6 mL
We need this volume in L, in order to reach molarity:
1318.6 mL . 1L / 1000mL = 1.3186 L ≅ 1.32L
Molarity (mol/L) → 8.01 mol / 1.32L = 6.07M
Hydroxylamine in water: HONH₂(aq) + H₂O(l) ⇄ HONH₃⁺(aq) + OH⁻(aq).
Hydroxylammonium nitrate in water: HONH₃NO₃(aq) → OHNH₃⁺(aq) + NO₃⁻(aq).
1) with positive hydrogen ions (protons) react base and gives weak conjugate acid:
H⁺(aq) + HONH₂(aq) ⇄ HONH₃⁺(aq).
2) with hydroxide anions react acid and produce weak base and weak electrolyte water:
HONH₃⁺(aq) + OH⁻(aq) ⇄ HONH₂(aq) + H₂O(l).
Molality is one way of expressing concentration for solutions. It has units of moles of solute per kg of solvent. From the given values, we easily calculate for the moles of solute by multiplying the mass of solvent to the molality. We do as follows:
moles solute = 0.3 (10) = 3 mol solute
