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Determine the equivalent weights of the following: K_2SO_4

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To determine the equivalent weight of K₂SO₄, we will follow these steps: ### Step 1: Determine the Molecular Weight of K₂SO₄ First, we need to calculate the molecular weight of K₂SO₄ (potassium sulfate). - Potassium (K) has an atomic weight of approximately 39 g/mol. - Sulfur (S) has an atomic weight of approximately 32 g/mol. - Oxygen (O) has an atomic weight of approximately 16 g/mol. The formula for K₂SO₄ consists of: - 2 Potassium atoms: 2 × 39 g/mol = 78 g/mol - 1 Sulfur atom: 1 × 32 g/mol = 32 g/mol - 4 Oxygen atoms: 4 × 16 g/mol = 64 g/mol Now, we can sum these values to find the total molecular weight: \[ \text{Molecular weight of K}_2\text{SO}_4 = 78 + 32 + 64 = 174 \text{ g/mol} \] ### Step 2: Determine the Valency of K₂SO₄ Next, we need to determine the valency of K₂SO₄. K₂SO₄ dissociates in solution into: - 2 K⁺ ions (each with a charge of +1) - 1 SO₄²⁻ ion (with a charge of -2) To find the valency, we consider the total positive and negative charges: - Total positive charge = 2 × (+1) = +2 - Total negative charge = -2 The valency is determined by the magnitude of the charge, which in this case is 2. ### Step 3: Calculate the Equivalent Weight The equivalent weight can be calculated using the formula: \[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Valency}} \] Substituting the values we found: \[ \text{Equivalent weight} = \frac{174 \text{ g/mol}}{2} = 87 \text{ g/equiv} \] ### Final Answer The equivalent weight of K₂SO₄ is **87 g/equiv**. ---
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RESONANCE ENGLISH-EQUIVALENT CONCEPT & TITRATIONS-Exercise -1 (Part-I )
  1. Determine the equivalent weight of Nacl

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  2. Determine the equivalent weights of the following: K2SO4

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  3. 1.12 litre dry chlorine gas at STP was passed over a heated metal when...

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  4. A mixture of CuS (molecular weight = M(1)) and Cu(2)S (molecular weigh...

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  5. Determine the equivalent weight of the following oxidising and reducin...

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  6. 0.98 g of the metal sulphate was dissolved in water and excess of bari...

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  7. A dilute solution of H(2)SO(4) is made by adding 5 mL of 3N H(2)SO(4) ...

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  8. What volume at NTP of gaseous ammonia will be required to be passed in...

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  9. 1.60 g of a metal A and 0.96 g of a metal B when treated with excess ...

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  10. It requires 40.05 ml of 1MCe^(4+) to titrate 20ml of 1 M Sn ^(2+) to S...

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  11. 25 mL of a solution of Fe^(2+) ions was titrated with a solution of th...

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  12. How many ml of 0.3M K2 Cr2 O7(acidic) is required for complete oxidati...

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  13. A mixture of H(2)SO(4) and H(2)C(2)O(4) (oxalic acid ) and some inert ...

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  14. A mixture containing As(2)O(3) and As(2)O(3) requried 20 " mL of " 0.0...

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  15. 20mL of H(2)O(2) after acidification with dilute H(2)SO(4) required 30...

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  16. A 100 mL sample of water was treated to convert any iron present to Fe...

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  17. What causes the temporary and permanent hardness of water ?

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  18. A driver having a definite reaction time is capable of stopping his ca...

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