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2 years ago

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In trying to control fall armyworms in crops, an Agriculture extension officer applied cypermethrin which was prepared by dissol

ving 200g of the cypermethrin , C22H19Cl2NO3 in 1000g of water H2O . Calculate the mole fraction of cypermethrin in the solution.

1 answer:

sveta [45]2 years ago

8 0

Answer:

Mole fraction for C₂₂H₁₉Cl₂NO₃ = 0.0086

Explanation:

Mole fraction remains a sort of concentration. It indicates:

moles of solute / (moles of solute + moles of solvent)

Moles of solute / Total moles.

Solute: Cypermethrin → C₂₂H₁₉Cl₂NO₃

Solvent: Water (PM = 18g/mol)

We calculate moles from solvent: 1000g /18 g/mol = 55.5 moles

We calculate PM for C₂₂H₁₉Cl₂NO₃

12g/mol . 22 + 1g/mol . 19 + 35.45 g/mol . 2+ 14g/mol + 16g/mol . 3 = 416 g/m

Moles of solute: 200 g / 416g/mol = 0.481 moles

Total moles: 0.481 + 55.5 = 55.98 moles

Mole fraction for C₂₂H₁₉Cl₂NO₃ = 0.481 moles / 55.98 moles = 0.0086

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He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o

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Answer : The activation energy of the reaction is, $17.285\times 10^4kJ/mole$

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The relation between the rate constant the activation energy is,

$\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = initial rate constant = $4.55\times 10^{-5}L/mole\text{ s}$

$K_2$ = final rate constant = $8.75\times 10^{-3}L/mole\text{ s}$

$T_1$ = initial temperature = $195^oC=273+195=468K$

$T_2$ = final temperature = $258^oC=273+258=531K$

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Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]$

$Ea=17.285\times 10^4kJ/mole$

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$K_{sp} = [Ag^{+}][Cl^{-}]$

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Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

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So, for this cell reaction the number of moles of electrons transferred are n = 1.

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$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

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