Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻
1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions
the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol
NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol
total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³
Answer:
30.12 g/m is the molar mass of the compound
Explanation:
Freezing point depression to solve this. The formula for the colligative property is:
ΔT = Kf . m
ΔT = T° freezing pure solvent - T° freezing solution
Kf = Cryoscopic constant
m = molality (mol/kg)
T° freezing pure benzene: 5.5°C
(5.5°C - 1.02°C) = 5.12 °C/m . m
4.48°C = 5.12 °C/m . m
4.48°C / 5.12 °C/m = m → 0.875 mol/kg
Mol = mass / molar mas
Molality = mol /kg
Let's find out the molar mass, with this equation:
(6.59 g / Molar mass) / 0.250 kg = 0.875 mol/kg
6.59 g / molar mass = 0.875 mol/kg . 0.250 kg
6.59 g / molar mass = 0.21875 mol
6.59 g / 0.21875 mol = molar mass → 30.12 g/m
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