The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:
0.115 mol I₂
1 - 0.115 = 0.885 mol CH₂Cl₂
We need moles of solute, which we have, and must convert our moles of solvent to kg:
0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂
We can now calculate the molality:
m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂
The molality of the iodine solution is 1.53.
Answer:
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Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
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