Answer :
The molecular weight of limonene and the relative error can be calculated using the experimental weights given and the Ideal Gas Law. The relative error is determined from the theoretical molecular weight.
The question provides the information that the weight of beaker₁ is 98.1 g and the weight of beaker₁ along with limonene is 113.98 g. This tells us that the weight of limonene is 113.98g - 98.1g = 15.88g.
Given that the atmospheric pressure is 1 atm and the vapor pressure of water at the given temperature is 0.719 atm, our pressure for limonene becomes 1 - 0.719 = 0.281 atm.
The given weight of water is 55.5 g - 67.20 g = -11.7 g (since weight cannot be negative, the experiment might have a mistake). Assuming an error and the true weight is 11.7 g, the moles of water would be 11.7/18.02 = 0.65 moles.
We use ideal gas law conversion: 0.281 atm = moles of limonene*0.08206*(92+273)/15.88g; solve to find the number of moles, and then finally the molecular weight of limonene.
Relative error can be calculated using the theoretical molecular weight which is specified as 136.24 g/mol. The formula for relative error is ((experimental value - theoretical value) / theoretical value) * 100%. The substituted values will give us the relative error.
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