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If 100 mL of 0.6 N H(2)SO(4) and 200 mL ...

If `100 mL` of `0.6 N H_(2)SO_(4)` and `200 mL` of `0.3 N HCl` are mixed together, the normality of the resulting solution will be

A

`(2N)/(5)`

B

`(N)/(10)`

C

`(N)/(5)`

D

`(N)/(20)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the normality of the resulting solution when mixing `100 mL` of `0.6 N H₂SO₄` and `200 mL` of `0.3 N HCl`, we can follow these steps: ### Step 1: Identify the given values - For H₂SO₄: - Normality (N₁) = `0.6 N` - Volume (V₁) = `100 mL` - For HCl: - Normality (N₂) = `0.3 N` - Volume (V₂) = `200 mL` ### Step 2: Calculate the total volume of the mixed solution The total volume (V₃) of the resulting solution is the sum of the volumes of the two solutions: \[ V₃ = V₁ + V₂ = 100 \, \text{mL} + 200 \, \text{mL} = 300 \, \text{mL} \] ### Step 3: Use the formula for normality of the resulting solution The normality of the resulting solution (N₃) can be calculated using the formula: \[ N₃ = \frac{N₁ \cdot V₁ + N₂ \cdot V₂}{V₃} \] ### Step 4: Substitute the values into the formula Substituting the known values into the formula: \[ N₃ = \frac{(0.6 \, N \cdot 100 \, mL) + (0.3 \, N \cdot 200 \, mL)}{300 \, mL} \] ### Step 5: Calculate the contributions from each solution Calculating each term: - Contribution from H₂SO₄: \[ 0.6 \, N \cdot 100 \, mL = 60 \] - Contribution from HCl: \[ 0.3 \, N \cdot 200 \, mL = 60 \] ### Step 6: Add the contributions and divide by the total volume Now, adding the contributions: \[ N₃ = \frac{60 + 60}{300} = \frac{120}{300} \] ### Step 7: Simplify the fraction Simplifying the fraction: \[ N₃ = \frac{2}{5} \, N \] ### Conclusion The normality of the resulting solution is: \[ N₃ = 0.4 \, N \]
To find the normality of the resulting solution when mixing `100 mL` of `0.6 N H₂SO₄` and `200 mL` of `0.3 N HCl`, we can follow these steps: ### Step 1: Identify the given values - For H₂SO₄: - Normality (N₁) = `0.6 N` - Volume (V₁) = `100 mL` - For HCl: ...
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