How many grams of sulfur (S8S8) must be dissolved in 100.0 g naphthalene (C10H8C10H8) to make a 0.11 mm solution

Respuesta :

Answer:

5.64x10⁻³ g of S₈S₈

Explanation:

We consider mm as milimolal

Milimolal = molal . 1000

Molal are the moles of solute contained in 1kg of solvent.

Solute → Sulfur (S₈S₈)

Solvent → naphthalene

0.11 mm / 1000 = 1.1x10⁻⁴ molal

moles of solute / kg of solvent = 1.1x10⁻⁴ molal

moles of solute / 0.1 kg of solvent = 1.1x10⁻⁴ molal

moles of solute = 1.1x10⁻⁴ m/kg .  0.1kg → 1.1ₓ10⁻⁵ moles

Molar mass S₈S₈ = 512.96 g/m

1.1ₓ10⁻⁵ mol . 512.96 g/m = 5.64x10⁻³ g