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The tablets were crushed, and 4.9993 g of the powder was transferred to a beaker and reacted with HCl. After filtration, the filtrate was transferred to a 100-mL volumetric flask and diluted with water. 20.00 mL of this stock solution were combined with 0.2 M Na3PO4. The resulting precipitate weighed 0.3451 g after drying. Calculate the moles of BiPO4 precipitated, the moles of Bi3 in the stock solution, and the mass of BSS per tablet.

Answer :

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Answer:

Explanation:

From the information given:

Mass of BiPO₄ = 0.3451 g

Number of moles of BiPO₄ = [tex]0.3451 \ g \ BiPO_4 \times \dfrac{1 \ mol \ BiPO_4}{303.95 \ g \ BiPO_4}[/tex]

[tex]= 0.001135 \ mol[/tex]

The number of moles of Bi³⁺ in 20.00 mL is:[tex]= 0.001135 \ mol \ BiPO_4 \times \dfrac{1 \ mol \ of \ Bi^{3+}}{1 \ mol \ BiPO_4}[/tex]

=  0.001135 mol of Bi³⁺

The number of moles of Bi³⁺ in 100 mL stock solution

[tex]= 0.001135 \ mol \ Bi^{3+} \times \dfrac{100 \ mL}{20.0 \ mL}[/tex]

[tex]= 0.005675 \ mol[/tex]

Mass of BSS in 4.9993 g tablets

[tex]m = 0.005675 \ mol \ Bi^{3+} \times \dfrac{1 \ mol \ BSS}{1 \ mol \Bi^{3+}} \times \dfrac{362.1 \ g \ BSS}{1 \ mol \ BSS}[/tex]

m = 2.055 g BSS

Mass of BSS in 5.0103 g (5 tables)

[tex]m = 2.055 g \ BSS \times \dfrac{5.0103 \ g}{4.9993 \ g}[/tex]

= 2.06 g

∴

The mass of BSS per tablet is [tex]=\dfrac{2.06 \ g}{5 \ tablet}[/tex]

= 0.412 g BSS/ tablet

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