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ser-zykov [4K]
3 years ago
7

Pure nitrobenzene freezes at 5.67 C. When 1.0g of ethanol (C2H6O) is mixed with 20.0g nitrobenzene, the freeze point drops to –0

.53 C. What is the freezing-point depression constant (Kf) of nitrobenzene?
Chemistry
1 answer:
sukhopar [10]3 years ago
5 0

Answer: The freezing-point depression constant (Kf) of nitrobenzene is 5.7^0C/m

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(5.67-(-0.53)^0C=6.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte nitrobenzene)

K_f = freezing point constant = ?

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent = 20 g = 0.02 kg

mass of solute (ethanol) = 1.0 g

Molar mass of ethanol = 46 g/mol

6.2=1\times K_f\times \frac{1.0g}{46g/mol\times 0.02kg}

K_f=5.7^0C/m

Thus freezing-point depression constant (Kf) of nitrobenzene is 5.7^0C/m

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Answer:

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Step 1: <u>Calculate Moles of Starting Materials:</u>

Moles of HCl:

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Step 2: <u>Find out Limiting reagent as:</u>

According to balance chemical equation.

              1 mole of HCl reacts with  =  1 mole of AgC₂H₃O₂

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         1.99 moles of HCl will react with  =  X moles of AgC₂H₃O₂

Solving for X,

                     X =  1.99 mol × 1 mol / 1 mol

                     X =  1.99 mol of AgC₂H₃O₂

Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.

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