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Sagot :
Percentage by weight of the NaNO₃, is given by the ratio of the mass of NaNO₃ to the mass of the sample
The percentage by weight of NaNO₃ in the sample is approximately 30.713 %
Reason:
Given parameters are;
Mass of sample of NaBr, NaI, and NaNO₃ mixture = 0.8612 g
Mass of AgNO₃ and AgI formed = 1.0186
Mass of AgCl formed = 0.7125
Required:
Percentage by weight of NaNO₃ in the sample
Solution:
Let A represent NaBr, let B represent NaI, and let C represent NaNO₃, we have;
Equation 1 A + B + C = 0.8612
The addition of AgNO₃, precipitates 1.0186 g mixture of AgBr and AgI, therefore, we have;
[tex]A \cdot\left(\dfrac{Molar \ mass \ of \ AgBr}{Molar \ mass \ of \ NaBr} \right) + B \cdot \left(\dfrac{Molar \ mass \ of \ AgI}{Molar \ mass \ of \ NaI} \right) = 1.0186[/tex]
Molar mass of AgBr = 187.77 g/mol
Molar mass of NaBr = 102.894 g/mol
Molar mass of AgI = 234.77 g/mol
Molar mass of NaI = 149.89 g/mol
Which gives;
Equation 2 [tex]A \cdot\left(\dfrac{187.77}{102.894} \right) + B \cdot \left(\dfrac{234.77}{149.89} \right) = 1.0186[/tex]
The forming of the precipitate by passing the compounds through a stream of chlorine gives;
[tex]A \cdot\left(\dfrac{Molar \ mass \ of \ AgCl}{Molar \ mass \ of \ NaBr} \right) + B \cdot \left(\dfrac{Molar \ mass \ of \ AgCl}{Molar \ mass \ of \ NaI} \right) = 0.7125[/tex]
Molar mass of AgCl = 143.32 g/mol
Therefore, we get;
Equation 3 [tex]A \cdot\left(\dfrac{143.32}{102.894} \right) + B \cdot \left(\dfrac{143.32}{149.89} \right) = 0.7125[/tex]
Solving equation (2), and (3) gives;
A ≈ 0.3252 g, B ≈ 0.2715 g
Plugging in the values of A, and B in equation (1), gives;
0.3252 + 0.2715 + C = 0.8612
C = 0.8612 - (0.3252 + 0.2715) ≈ 0.2645
The mass of NaNO₃ in the mixture, C ≈ 0.2645
[tex]The \ percent \ by \ weight \ of \ NaNO_3 = \dfrac{0.2645}{0.8612} \times 100 \approx \underline{30.713 \%}[/tex]
Learn more here:
https://brainly.com/question/14674489
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