Answer to Question #169772 in General Chemistry for Zikrullah

Question

A solution of ammonium sulphate fertilizer was to be applied to the soil before planting some seeds. A given volume of water was put in a large container and 3.68kg of fertilizer was dissolved in it. On analysis, 0.28dm³ of 0.35moldm^-3 barium chloride was required to precipitate all the sulphate in 0.20dm³ of solution. What was the volume of the solution in the container? {H=1, N=14, O=16, S=32}

Answer 1

Q169772

 

A solution of ammonium sulphate fertilizer was to be applied to the soil before planting some seeds. A given volume of water was put in a large container and 3.68kg of fertilizer was dissolved in it. On analysis, 0.28dm³ of 0.35moldm^-3 barium chloride was required to precipitate all the sulphate in 0.20dm³ of solution. What was the volume of the solution in the container? {H=1, N=14, O=16, S=32}

Solution:

A 0.20 dm³ sample of ammonium sulphate solution is taken and is analyzed for the sulphate content.

We will first find the concentration of sulphate in the sample of ammonium sulphate solution.

The reaction of sulphate and barium chloride is given as

SO₄^2- (aq) + BaCl₂ (aq) ===> BaSO₄ (s) + 2 Cl^- (aq)

Step 1 : Find the moles of BaCl₂ required for analyzing the given sample of ammonium sulphate.

$Molarity = \frac{moles \space of \space BaCl_2}{volume \space of \space BaCl_2 \space in \space 'L' }$

1dm³ = 1 L, So 0.28 dm³ = 0.28 L.

Also 0.35 mol dm³ can be written as 0.35 mol/ L .

plug this information in the formula we have

$0.35mol/L = \frac{moles \space of \space BaCl_2}{0.28L }$

multiply both the side by 0.28 L, we have

moles of BaCl₂ = 0.35mol/L * 0.28 L = 0.098 mol of BaCl₂ .

Step 2 : To find the moles of sulphate present in the given sample of ammonium sulphate.

The reaction of sulphate and barium chloride is given as.

1 SO₄^2- (aq) + BaCl₂ (aq) ===> BaSO₄ (s) + 2 Cl^- (aq)

The mol to mol ratio of SO₄^2- and BaCl₂ in this reaction is 1 :1.

So the moles of sulphate present in the sample will too be equal to 0.098 mol SO₄^2- .

Step 3 : To find the moles of ammonium sulphate in the sample that will give 0.098 moles of

SO₄^2- .

The dissociation of ammonium sulhpate is given as

1 (NH₄)₂ SO₄ (aq) ===> 2 NH₄ ⁺(aq) + 1 SO₄ ^2- (aq)

the mole to mole ratio of (NH₄)₂ SO₄ and SO₄ ^2- is 1 :1.

so moles of ammonium sulphate present in the sample will too be equal to 0.098 moles.

Step 4 : To find the concentration of (NH₄)₂ SO₄ in the sample.

the volume of (NH₄)₂ SO₄ taken for the analysis is 0.20 dm³.

which in 'L' is 0.20 L.

plug 0.098 mol (NH₄)₂ SO₄ and volume = 0.20L, and find the molarity of (NH₄)₂ SO₄ in the sample.

$Molarity\space of \space (NH_4)_2 SO_4 = \frac{moles\space of\space (NH_4)_2 SO_4 }{volume \space of \space (NH_4)_2 SO_4\space solution\space in\space 'L' }$

$Molarity\space of \space (NH_4)_2 SO_4 = \frac{0.098 \space mol\space of\space (NH_4)_2 SO_4 }{0.20\space L }$

Molarity of ammonium sulphate = 0.49mol/L

The concentration of ammonium sulphate in the large container will be same as that of the concentration of ammonium sulphate in the sample taken from the large container.

Step 5: To convert the mass of fertilizer (ammonium sulphate ) to grams.

3.68kg of fertilizer was dissolved in the large reservoir.

Convert 3.68 kg to moles by using molar mass of ammonium sulphate.

molar mass of (NH₄)₂ SO₄

= 2 * atomic mass of N + 8 * atomic mass of H + 1 * atomic mass of S + 4 * atomic mass of O

= 2 * 14g/mol + 8 * 1g/mol + 1 * 32 g/mol + 4 * 16 g/mol

= 28 + 8 + 32 + 64

= 132 g/mol

3.68 kg in grams = $3.68 kg * \frac{1000g }{1kg } = 3680 g$

moles of (NH₄)₂ SO₄ = $3680 g * \frac{1\space mol }{132 \space g } = 27.88 \space moles \space of \space (NH_4)_2 SO_4$

Step 6 : To find the volume of water in the large container.

Now we have total moles of ammonium sulphate in container = 27.88 moles

concentration of ammonium sulphate in the container = 0.49 mol/L

plug this in the molarity formula again and find the volume of cotainer.

$Molarity\space of\space (NH_4)_2 SO_4 = \frac{moles of (NH_4)_2 SO_4 }{volume\space of\space container\space in\space 'L' }$

$0.49mol/L = \frac{27.88\space moles }{volume\space of\space container\space in\space 'L' }$

arranging this equation for volume, we have

volume of container $= \frac{27.88 \space moles }{0.49\space mol/L } = 56.90L$

In the given quesiton the concentration and volume are given in 2 significant figure,

so our final answer must also be in 2 significant figure.

Hence the volume of the container is 57 Liters or 57 dm³ .