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3 years ago

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Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0

.771 g of biphenyl (C12H10) in 25.0 g of cyclohexane?

1 answer:

Elza [17]3 years ago

4 0

Answer: $2.49^0C$

Explanation:

Depression in freezing point is:

$T_f^0-T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of solvent (cyclohexane) = $6.50^oC$

$k_f$ = freezing point constant of solvent (cyclohexane) = $20.0^oC/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute (biphenyl) = 0.771 g

$w_1$ = mass of solvent (cyclohexane) = 25.0 g

$M_2$ = molar mass of solute (biphenyl) =

Now put all the given values in the above formula, we get:

$(6.50-T_f)^oC=1\times (20.0^oC/m)\times \frac{(0.771g)\times 1000}{154\times (25.0g)}$

$(6.50-T_f)^oC=4.01$

$T_f=2.49^0C$

Therefore, the freezing point of a solution made by dissolving 0.771 g of biphenyl in 25.0 g of cyclohexane is $2.49^0C$

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Answer:

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Explanation:

From the question we are told that

The equation is $2 Cr + 3Pb^{2+} \to 3Pb + 2Cr^{3+}$

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Generally at the cathode

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