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A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS₂ (Kb = 2.34 °C/m). If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide (assuming ideal solution behavior)?

a) S₈
b) S₄
c) S₂
d) S₁₆

1 Answer

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Final answer:

The formula of the sulfur molecule dissolved in carbon disulfide, given the boiling point elevation of 0.107 °C, is determined to be S8. Therefore, the answer is a) S₈.

Step-by-step explanation:

To determine the formula of a sulfur molecule in carbon disulfide, we need to use the boiling point elevation formula ΔTb = Kb * m, where ΔTb is the boiling point elevation, Kb is the ebullioscopic constant of the solvent (carbon disulfide), and m is the molality of the solution. First, we calculate the molality (m) of the sulfur in the solution:

m = moles of solute / kilograms of solvent

Given the boiling point elevation (ΔTb) is 0.107 °C and the Kb for carbon disulfide (CS₂) is 2.34 °C/m, we have:

0.107 °C = 2.34 °C/m * m

m = 0.107 °C / 2.34 °C/m = 0.0457 m

The number of moles of sulfur is calculated by:

moles of sulfur = molality (m) * kilograms of solvent

moles of sulfur = 0.0457 m * 0.0178 kg = 0.00081346 mol

Next, we calculate the molar mass (M) of sulfur:

M = mass of solute / moles of solute

M = 0.210 g / 0.00081346 mol = 258.2 g/mol

The molar mass of atomic sulfur (S) is 32 g/mol, so we divide the molar mass of sulfur in the solution by the molar mass of atomic sulfur to find the number of sulfur atoms per molecule:

258.2 g/mol / 32 g/mol = 8.07 ≈ 8

Since we can only have whole atoms in a molecule, the sulfur molecule is most likely to have 8 atoms, forming an S8 molecule. Therefore, the answer is a) S₈.

User Narendra Kothule
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