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100 mL of a sample of hard water require...

`100 mL` of a sample of hard water requires `25.1 mL` of `0.02 N H_(2)SO_(4)` for complete reaction, The hardness of water ( density `1g//mL)` is:

A

`200"ppm"`

B

`250"ppm"`

C

`251"ppm"`

D

`258"ppm"`

Text Solution

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The correct Answer is:
To find the hardness of water in ppm, we can follow these steps: ### Step 1: Calculate the number of milliequivalents of H₂SO₄ used To find the milliequivalents of H₂SO₄, we use the formula: \[ \text{Milliequivalents} = \text{Volume (mL)} \times \text{Normality (N)} \] Given: - Volume of H₂SO₄ = 25.1 mL - Normality of H₂SO₄ = 0.02 N Calculating: \[ \text{Milliequivalents} = 25.1 \, \text{mL} \times 0.02 \, \text{N} = 0.502 \, \text{mEq} \] ### Step 2: Relate the milliequivalents of H₂SO₄ to the hardness of water The hardness of water is often expressed in terms of CaCO₃. The equivalent weight of CaCO₃ is calculated as follows: \[ \text{Equivalent weight of CaCO₃} = \frac{\text{Molar mass of CaCO₃}}{2} = \frac{100 \, \text{g/mol}}{2} = 50 \, \text{g/equiv} \] ### Step 3: Calculate the weight of CaCO₃ equivalent to the milliequivalents Using the milliequivalents calculated in Step 1, we can find the weight of CaCO₃: \[ \text{Weight of CaCO₃} = \text{Milliequivalents} \times \text{Equivalent weight} = 0.502 \, \text{mEq} \times 50 \, \text{g/equiv} = 25.1 \, \text{g} \] ### Step 4: Calculate the total weight of the water sample Given that the density of water is 1 g/mL, the weight of the 100 mL water sample is: \[ \text{Weight of water sample} = \text{Volume} \times \text{Density} = 100 \, \text{mL} \times 1 \, \text{g/mL} = 100 \, \text{g} \] ### Step 5: Calculate the hardness in ppm To convert the weight of CaCO₃ to ppm, we use the formula: \[ \text{Hardness (ppm)} = \left( \frac{\text{Weight of CaCO₃}}{\text{Weight of water sample}} \right) \times 10^6 \] Calculating: \[ \text{Hardness (ppm)} = \left( \frac{25.1 \, \text{g}}{100 \, \text{g}} \right) \times 10^6 = 251000 \, \text{ppm} \] ### Step 6: Final Adjustment Since we are looking for the hardness in ppm, we can directly state that the hardness of water is: \[ \text{Hardness} = 251 \, \text{ppm} \] ### Conclusion The hardness of the water sample is **251 ppm**. ---

To find the hardness of water in ppm, we can follow these steps: ### Step 1: Calculate the number of milliequivalents of H₂SO₄ used To find the milliequivalents of H₂SO₄, we use the formula: \[ \text{Milliequivalents} = \text{Volume (mL)} \times \text{Normality (N)} \] ...
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P BAHADUR-MOLE AND EQUIVALENT CONCEPT-Exercise (3A) Objective problems:
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  2. 0.05 moles of NaHCO(3) will react with how many equivalent of Mg(OH)(2...

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  3. 0.078 g Al(OH)(3). is dehydrated to Al(2)O(3). The Al(2)O(3) so obtain...

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  4. Assuming 100% ionisation, the solution having highest normality is:

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  5. 100 mL of a sample of hard water requires 25.1 mL of 0.02 N H(2)SO(4) ...

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  6. The equivalent weight of potash alum (K(2)SO(4).Al(2)(SO(4))(3).24H(...

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  7. Vapour density of a metal chloride is 6.6. Its oxide contains 53% meta...

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  8. 100 mL of 0.1 M solution of H(2)SO(4) is used to prepare 0.05N solutio...

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  9. Which does not change on dilution?

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  10. 20 mL of 0.1M H(3)BO(3) solution on complete netralisation requires …....

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  11. Volume strength of H(2)O(2) labelled is 10 vol. What is normality of H...

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  12. Find the weight of H(2) SO(4) in 1200 mL of a solution of 0.2 N streng...

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  13. Calculate the volume of 0.5 M H2SO4 required to dissolve 0.5 g of copp...

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  14. 1 g of calcium was burnt in excess of O(2) and the oxide was dissolved...

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  15. Number of H^(+) ions in 100 mL of 0.001 M H(2)SO(4) is:

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  16. 3 g of an oxide of a metal is converted completely to 5 g chloride. Eq...

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  17. V(1) mL of NaOH of normality X and V(2) mL of Ba(OH)(2) of mormality Y...

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  18. Weight of oxygen in Fe(2)O(3) and FeO in the simple ratio for the same...

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  19. An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 m...

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  20. 0.63 g of diabasic acid was dissolved in water. The volume of the solu...

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