Use the References to access important valnes if needed for this question. The n

Question

Use the References to access important valnes if needed for this question. The normal freezing point of water (H2O) is 0.000°C and its Kfp value is 1.86 °Cm In a laboratory experiment, students synthesized a new compound and found that when 14.50 grams of the compound were dissolved in 289.0 grams of water, the solution began to freeze at -1.503 C. The compound was also found to be nonvolatile and a nonelectrolyte What is the molecular weight they determined for this compound? g mol Submit Answer

Explanation / Answer

we have below equation to be used:

delta Tf = Kf*mb

1.503 = 1.86 *mb

mb= 0.8081 molal

mass of solvent = 289 g

= 0.289 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

number of mol,

n = Molality * mass of solvent in Kg

= (0.8081 mol/Kg)*(0.289 Kg)

= 0.2335 mol

mass of solute = 14.5 g

we have below equation to be used:

number of mol = mass / molar mass

0.2335 mol = (14.5 g)/molar mass

molar mass = 62.1 g/mol

Answer: 62.1 g/mol

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