Please Help ALL parts A, B, and C b.) Consider the following data for the freezi
ID: 556403 • Letter: P
Question
Please Help ALL parts A, B, and C
b.) Consider the following data for the freezing point depressions of dilute butanol-water solutions. molality ATI Mole Fraction Butanol Mole Fraction Water Activity of Water 0.004134 0.007669 0.019330 0.03558 0.02420 0.04452 0.03982 0.07300 0.05206 0.09505 0.08087 0.14679 0.09786 0.17680 a.) Convert the molality of the butanol into mole fractions of butanol and water. b.) Calculate the activity of the water in each of the solutions. c.) Curve fit the ratio of the natural log of activity of water to the mole fraction to the square of the butanol mole fraction. That is, fit the following... In (Owater /xwater) = A • Xbutanol Determine the value of the coefficient, A.Explanation / Answer
Answer:
As we know that
Molality = (moles of solute/ given mass of solvent in kg)
given mass of solute = M1, the mass of solvent = M2 and molecular mass of solute = 74g
So, we can also write molality = (given mass of solute/ molecular mass of solute) / given mass of solvent in kg
Case 1 Molality = M1/M2*74 = 0.004134
=> M1/M2 = 0.004134/74
So, we know that mole fraction of solute = mole of solute/total moles of solution ( mole of solute = m1, mole of solvent = m2)
Therefor mole fraction of solute (x1) = (m1)/m1+m2)
So, x1 = (M1/ 74)/ (M1/74 + M2/18)
x1 = 1 / (1 + (M2/M1) *(74/18))
x1 = 1 / ( 1 + 4.1 * (1/0.00006)) ( M1/M2 = 0.004134/74)
x1 = 1 / 68334.306 = 1 * 10^-5
mole fraction of solvent (x2)= 1 - x1 = 1 - 0.00001
x2 = 0.9999
follow same process for all other solution of given molality
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b) Activity of water is defined by freezing point depression constant
therefor
we know that depression in freezing point relation
delta Tf = Kf * m
K f = freezing point depression constant, m = molality
therefor for case 1 Tf = 0.007669
=> 0.007669 = K f * 0.004134
=> K f = 0.53846 K kg/ mol = activity of water
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c) given relation
In ( awater / x2 ) = A. x12
In (0.53846 K kg/ mol / 0.9999) = A * 1 * 10^-10
= > A = In (0.53851) / 1 * 10^-10
=> A = - 0.6189 *10^10
the same process you can use for calculation of mole fraction, the activity of water and value of 'A' of a component of given solution.
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