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Number of milli equivalents of solute in...

Number of milli equivalents of solute in 0.5 litres of 0.2 N solution is

A

10

B

1

C

100

D

1000

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The correct Answer is:
To find the number of milli equivalents of solute in a 0.5 liters of 0.2 N solution, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Definition of Normality (N)**: Normality (N) is defined as the number of equivalents of solute per liter of solution. The formula is: \[ N = \frac{\text{Number of equivalents}}{\text{Volume of solution in liters}} \] 2. **Convert Normality to Milli Equivalents**: Since we need to find the number of milli equivalents, we can modify the formula to: \[ N = \frac{\text{Number of milli equivalents}}{\text{Volume of solution in milliliters}} \] 3. **Rearranging the Formula**: Rearranging the formula to find the number of milli equivalents: \[ \text{Number of milli equivalents} = N \times \text{Volume of solution in milliliters} \] 4. **Substituting Given Values**: We are given: - Normality (N) = 0.2 N - Volume of solution = 0.5 liters First, convert the volume from liters to milliliters: \[ 0.5 \text{ liters} = 0.5 \times 1000 = 500 \text{ milliliters} \] 5. **Calculating the Number of Milli Equivalents**: Now, substitute the values into the formula: \[ \text{Number of milli equivalents} = 0.2 \times 500 = 100 \] ### Final Answer: The number of milli equivalents of solute in 0.5 liters of 0.2 N solution is **100 milli equivalents**. ---

To find the number of milli equivalents of solute in a 0.5 liters of 0.2 N solution, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Definition of Normality (N)**: Normality (N) is defined as the number of equivalents of solute per liter of solution. The formula is: \[ N = \frac{\text{Number of equivalents}}{\text{Volume of solution in liters}} ...
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NARAYNA-SOLUTIONS & COLLIGATIVE PROPERTIES -EXERCISE : 2 (C.W)
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  3. Number of milli equivalents of solute in 0.5 litres of 0.2 N solution ...

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  4. Volume of 0.1 M K(2)Cr(2)O(7) required to oxidize 35 mL of 0.5 M FeSO(...

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  5. Weight of solute present in 500 ml 0.2 N-H(2)SO(4) solution is

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  6. Number of milli equivalents of solute present in 250 ml of 0.1 M oxali...

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  7. The mass of oxalic acid crystals H(2)C(2)O(4).2H(2)O required to prepa...

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  8. What is the volume (in litres) of 0.1 M H(2)SO(4) required to complete...

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  9. 3.42 g of a substance of molecular weight 342 is present in 250g of wa...

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  11. The mole fraction of NaCl in a solution containing 1 mole of NaCl in 1...

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  14. A gaseous mixture contains 4.0g of H(2) and 56.0g of N(2). The mole fr...

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  15. Three statements are given about mole fraction (i) Mole fraction o...

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  16. Henry 's law constant for the solubility of N(2) gas in water at 298 K...

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  17. H(2)S, a toxic gas with rotten egg like smell, is used for the qualita...

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  18. Henry's law constant for CO(2) in water is 1.67xx10^(8) Pa at 298 K. ...

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  19. A solution is obtained by dissolving 0.2 moles of urea in a litre wate...

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