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scZoUnD [109]
3 years ago
9

he electrical leads from the Constant Current System are damaged slightly, leading to a small "short-circuit" that results in 10

% of the measured current passing through the short circuit without going through the metal strips. The current displayed (and hence the value of the integral) are based on the total current, but only 90% of this actually leads to dissolution of metal. What effect would this have on your experimental value of Avogadro's Number? (Note that you complete your calculations assuming that 100% of the current is applied to dissolving the metal strip.)
Chemistry
1 answer:
VikaD [51]3 years ago
8 0

Answer:

The Avogadro's Number will reduce in value

Explanation:

From the question we are told that

The current pass through the short circuit is 0.1 *I

Here I is the total current

Generally Avogadro's Number is mathematically represented as

N_A =  \frac{Q}{ 2 *  n * e}

Here Q is the charge which is mathematically represented as

Q =  It

        n  is number of moles

        e is the charge on an electron

 So

       N_A =  \frac{I * t }{ 2 *  n * e}

Now when the current decreases from 100% to  90% due to the short circuit . this will also reduce  the no of moles of metal  dissolved hence from the equation we see that the Avogadro's Number will reduce

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<u>Explanation:</u>

We are given:

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Mass percentage of C_2H_2O = 15 %

So, mole fraction of C_2H_2O = 0.15

We know that:

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To calculate the average molecular mass of the mixture, we use the equation:

\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}

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m_i = molar masses of i-th species

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Putting values in above equation:

\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}

\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75

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