What is the molarity of 930 mL of LiCl solution that contains 139.9 g of LiCl?
O A. 3.5 mol/L
O B. 150 mol/L
hipotle and like
O C. 3.3 mol/L
O D. 3.5 x 10-3 mol/L

What Is The Molarity Of 930 ML Of LiCl Solution That Contains 139.9 G Of LiCl?O A. 3.5 Mol/LO B. 150

Answers

Answer 1

Answer:

3.5 mol/L

Explanation:

First convert the 139.9g of LiCl into moles of LiCl.  Don't do the sig fig rule until the very end.

Second, take 930mL of LiCl and divide it by 1000 to convert it to liters.

Third use the molarity formula which is M=mol/L and divide 0.930L by 3.300773877mol and you will get 3.549219223.  This is where the sig fig rule will come into play.  3.5mol/L is a reasonable answer for this problem.


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A 0.8612-g sample of a mixture of NaBr, NaI, and NaNO3 was analyzed by adding AgNO3 and precipitating a 1.0186-g mixture of AgBrand AgI. The precipitate was then heated in a stream of Cl2, convertingit to 0.7125 g of AgCl. Calculate the %w/w NaNO3 in the sample.

Answers

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:

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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]

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Molar mass of AgI = 234.77 g/mol

Molar mass of NaI = 149.89 g/mol

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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]

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Equation 3 [tex]A \cdot\left(\dfrac{143.32}{102.894} \right) + B \cdot \left(\dfrac{143.32}{149.89} \right) = 0.7125[/tex]

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Plugging in the values of A, and B in equation (1), gives;

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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]

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