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Chemical EquilibriumClass 11thChemistry NCERT Exemplar Solutions Class 12th Chapter EightChemistry

The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

 

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mi mathvariant="normal">X</mi> <mi mathvariant="normal">Y</mi> <mo>,</mo> <mn>2</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>5</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi mathvariant="normal">M</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">X</mi> <mi mathvariant="normal">Y</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> <mo>,</mo> <mn>1</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>9</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi mathvariant="normal">M</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">X</mi> <mi mathvariant="normal">Y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>,</mo> <mn>4</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>9</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi mathvariant="normal">M</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">X</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mi mathvariant="normal">Y</mi> <mo>,</mo> <mn>2</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>9</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi mathvariant="normal">M</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> </span></p>
5 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
R
7 months ago
Correct Option - 3
Detailed Solution:

X Y 2 → X + 2 + 2 y - K   sp   = X + 2 [ X ] 2 = 10 - 3 2 * 10 - 3 2 = 4 * 10 - 9

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Enthalpy of neutralisation of CH3COOH by NaOH is – 50.6 kJ/mol and the heat of neutralisation of a strong acid with NaOH is – 55.9 kJ/mol. The value of ΔH for the ionisation of CH3COOH is :

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CH3COOH + NaOH → CH3COONa + H2O

ΔH = –50.6 kJ/mol

NaOH + SA [HCl] → NaCl + H2O

ΔH = –55.9 kJ/mol

the value of ΔH for ionisation of CH3COOH

⇒ ΔH = +55.9 – 50.6

5.3 kJ/mol

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