ΔT = T(boiling solution) – T(water boiling)
ΔT = 104 – 100 = 4 ºC
"m = \\frac{\u0394T}{K_b} \\\\\n\nm = \\frac{4}{0.52} = 7.69 \\;mol\/kg \\\\\n\nm = \\frac{moles\\;solute}{kg\\;solvetnt}"
kg solvent = 0.05 kg
moles solvent "= m \\times kg\\;solvent"
moles solvent "= 7.69 \\times 0.05 = 0.3845 \\;mol"
"m_c = 25.0 \\;g \\\\\n\nn_c = 0.3845 \\;mol \\\\\n\nMM_c = \\frac{25.0}{0.3845} = 65.02 \\;g\/mol"
Answer: 65.02 g/mol
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