With explanation What is the entropy change in Joules when one mole of an ideal gas...

Question

Question

With explanation

Answer 1

Answer #1

ΔS = nRln(Vf/Vi)

= 1*8.314 ln 2 = 5.7628 j/k

a. 5.7628 j/k

a.

1 kg benzene no of moles = 1000/78.11 = 12.80

freezing point is the same because of no solute

b.

calculate moles of solute that is naphtalene = 10/128 = 0.078125

molality of solution = moles of naphthalene / moles of benzene = 0.078125/12.8 = 0.0061 molal

= 4.3 * 0.0061 = 0.0263⁰C[freezing point has reduced by this much due to addition of solute napthalene]

c.

calculate moles of solute that is biphenyl= 10/154 = 0.0649

molality of solution = moles of biphenyl / moles of benzene = 0.0649/12.8 = 0.00507 molal

= 4.3 * 0.00507 = 0.0218⁰C[freezing point has reduced by this much due to addition of solute biphenyl].

So answer is B. 1 kg benzene with 10 g of biphenyl where freezing point has reduced by 0.0263⁰C due to addition of solute napthalene

Answer 2

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