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The normality of a solution obtained by ...

The normality of a solution obtained by mixing 100 ml of 0.2 N HCl and 500 ml of 0.12 M `H_(2) SO_(4)` is -

A

0.233N

B

0.466N

C

0.116N

D

2.33 N

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The correct Answer is:
To find the normality of the solution obtained by mixing 100 ml of 0.2 N HCl and 500 ml of 0.12 M H₂SO₄, we can follow these steps: ### Step 1: Calculate the Normality of H₂SO₄ The normality (N) of a solution can be calculated from its molarity (M) using the formula: \[ N = n \times M \] where \( n \) is the number of replaceable H⁺ ions (the acidic strength). For H₂SO₄, it can donate 2 H⁺ ions, so: - \( n = 2 \) - \( M = 0.12 \, \text{M} \) Calculating the normality: \[ N_{H_2SO_4} = 2 \times 0.12 = 0.24 \, \text{N} \] ### Step 2: Calculate Milliequivalents of HCl Milliequivalents (meq) can be calculated using the formula: \[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (ml)} \] For HCl: - Normality \( = 0.2 \, \text{N} \) - Volume \( = 100 \, \text{ml} \) Calculating the milliequivalents: \[ \text{meq}_{HCl} = 0.2 \times 100 = 20 \, \text{meq} \] ### Step 3: Calculate Milliequivalents of H₂SO₄ Using the same formula for H₂SO₄: - Normality \( = 0.24 \, \text{N} \) - Volume \( = 500 \, \text{ml} \) Calculating the milliequivalents: \[ \text{meq}_{H_2SO_4} = 0.24 \times 500 = 120 \, \text{meq} \] ### Step 4: Calculate Total Milliequivalents Now, we can find the total milliequivalents by adding the milliequivalents of both acids: \[ \text{Total meq} = \text{meq}_{HCl} + \text{meq}_{H_2SO_4} = 20 + 120 = 140 \, \text{meq} \] ### Step 5: Calculate Total Volume of the Solution The total volume of the solution is the sum of the volumes of the two solutions: \[ \text{Total Volume} = 100 \, \text{ml} + 500 \, \text{ml} = 600 \, \text{ml} \] ### Step 6: Calculate Normality of the Mixed Solution Finally, we can calculate the normality of the mixed solution using the formula: \[ N = \frac{\text{Total meq}}{\text{Total Volume (ml)}} \] Calculating the normality: \[ N = \frac{140}{600} = 0.233 \, \text{N} \] ### Final Answer The normality of the solution obtained by mixing the two acids is approximately **0.233 N**. ---
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MOTION-OXIDATION & REDUCTION -OXIDATION NUMBER ( EXERCISE -1)
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  2. 100 c.c. of 0.5 N NaOH Solution is added to 10 c.c. opf 3N H(2)SO(4) s...

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  3. The normality of a solution obtained by mixing 100 ml of 0.2 N HCl and...

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  4. 0.115gm of sodium metal was dissolved in 500ml of the solution in dist...

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  5. How much water should be added to 200 c.c of seminormal solution of Na...

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  6. The amount of water to be added to 100 c.c. of 1 normal HCl solution ...

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  7. The amount of CuSO(4). 5H(2) O required to prepare 500 c.c. of 0.5 N s...

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  8. It 100 mL of 0.6 N H(2)SO(4) and 200 mL of 0.3 N HCl are mixed togethe...

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  9. How many gram of KMnO(4) are contained in 4 litres of 0.05N solution? ...

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  10. 25 mL HNO(3). If the volumes are mixed with 75 mL of 4.0 M HNO(3). If ...

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  11. IF 250 mL of 0.25 M NaCl solution is diluted with water to a vo...

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  12. 100ml of 0.3 N HCl is mixed with 200ml of 0.6 N H(2)SO(4). The final n...

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  13. What is the normlity of a 1 M solution of H(3)PO(4)?

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  14. 9.8 g of H(2)SO(4) is present in two litres of a solution. The molarit...

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  15. 10ml of N-HCl 20ml of N//2 H(2)SO(4) and 30 ml of N//3 HNO(3) are mi...

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  16. On dissolving 1 mole of each of the following acids in 1 litre water, ...

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  17. a(1) gm of the metal gave a(2) gm of its oxide. Hence equivalent weigh...

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  18. 5.6 g of a metal forms 12.7 g of metal chloride. Hence equivalent weig...

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  19. 0.5g of a metal on oxidation gave 0.7 g of its oxide. The equivalent w...

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  20. An element forms an oxide, in which the oxygen is 20% of the oxide by ...

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