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ANTONII [103]
3 years ago
11

What is the value of freezing point depression of a solution of 15.0g of NaCl in 250g of water?

Chemistry
1 answer:
kobusy [5.1K]3 years ago
7 0

Answer:

-3.82ºC is the freezing point of solution

Explanation:

We work with the Freezing point depression to solve the problem

ΔT = m . Kf . i

ΔT = Freezing point of pure solvent - freezing point of solution

Let's find out m, molality (moles of solute in 1kg of solvent)

15 g / 58.45 g/mol = 0.257 moles of NaCl

NaCl(s) → Na⁺ (aq) + Cl⁻(aq)

i = 2 (Van't Hoff factor, numbers of ions dissolved)

m = mol /kg → 0.257 mol / 0.250kg = 1.03 m

Kf = Cryoscopic constant → 1.86 ºC/m (pure, for water)

0ºC - Tºf = 1.03m . 1.86ºC/m . 2

Tºf  = -3.82ºC

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Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

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Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

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5 0
2 years ago
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il63 [147K]
The correct answer is Crystals.

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4 0
3 years ago
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