Home
Class 12
CHEMISTRY
50 mL of 10 N H(2)SO(4), 25 mL of 12 N H...

`50 mL` of `10 N H_(2)SO_(4), 25 mL` of `12 N HCI` and `40 mL` of `5N HNO_(3)` are mixed and the volume of the mixture is made 1000 mL by adding water. The normality of resulting solution will be

A

1 N

B

2 N

C

3 N

D

4 N

Text Solution

AI Generated Solution

The correct Answer is:
To find the normality of the resulting solution after mixing the given solutions and diluting to a total volume of 1000 mL, we can follow these steps: ### Step 1: Identify the normalities and volumes of the solutions - For `H2SO4`: Normality (N1) = 10 N, Volume (V1) = 50 mL - For `HCl`: Normality (N2) = 12 N, Volume (V2) = 25 mL - For `HNO3`: Normality (N3) = 5 N, Volume (V3) = 40 mL ### Step 2: Calculate the equivalent milliequivalents for each solution The equivalent milliequivalents (mEq) can be calculated using the formula: \[ \text{mEq} = \text{Normality} \times \text{Volume (in L)} \] 1. For `H2SO4`: \[ \text{mEq}_1 = 10 \, \text{N} \times 0.050 \, \text{L} = 0.5 \, \text{mEq} \] 2. For `HCl`: \[ \text{mEq}_2 = 12 \, \text{N} \times 0.025 \, \text{L} = 0.3 \, \text{mEq} \] 3. For `HNO3`: \[ \text{mEq}_3 = 5 \, \text{N} \times 0.040 \, \text{L} = 0.2 \, \text{mEq} \] ### Step 3: Sum the equivalent milliequivalents Now, we add all the equivalent milliequivalents together: \[ \text{Total mEq} = \text{mEq}_1 + \text{mEq}_2 + \text{mEq}_3 = 0.5 + 0.3 + 0.2 = 1.0 \, \text{mEq} \] ### Step 4: Calculate the normality of the final solution The total volume of the mixture after dilution is 1000 mL (or 1 L). The normality of the resulting solution (N4) can be calculated using the formula: \[ N4 = \frac{\text{Total mEq}}{\text{Total Volume (in L)}} \] \[ N4 = \frac{1.0 \, \text{mEq}}{1 \, \text{L}} = 1 \, \text{N} \] ### Conclusion The normality of the resulting solution is **1 N**.
Promotional Banner

Similar Questions

Explore conceptually related problems

5 mL of N-HCl , 20 mL of N//2 H_(2)SO_(4) and 30 mL of N//3 HNO_(3) are mixed together and the volume is made to 1 L . The normality of the resulting solution is

5 mL of N HCI, 20 mL of N/2 H_2SO_4 and 30 mL of N/3 HNO_3 are mixed together and the volume made to 1 litre. The normality of the resulting solution is:

10mL of an N HCI, 20mL of N/2 H_2 SO_4 and 30 mL of N/3 HNO_3 are mixed together and volume made to one litre. The normality of H^+ in the resulting solution is :

10mL of N-HCl, 20mL of N//2 H_(2)SO_(4) and 30mL N//3 HNO_(3) are mixed togeher and volume made to one litre. The normally of H^(+) in the resulting solution is:

20 ml of 1N HCl, 10 ml of 'N/2 H_2SO_4 and 30ml of 'N/3 HNO_3 are mixed together and volume made to 1000 ml. Find out the normality of 'H^+ ions in the resulting solution

1 mL of 0.1 N HCl is added to 999 mL solution of NaCl. The pH of the resulting solution will be :

20 mL N/2 HCI, 60 mL N/10 H_2SO_4 and 150 mL N/5 HNO_3 are mixed. Calculate the normality of the mixture of acids in solution.

50 mL of 1 M HCl, 100 mL of 0.5 M HNO_(3) , and x mL of 5 M H_(2)SO_(4) are mixed together and the total volume is made upto 1.0 L with water. 100 mL of this solution exactly neutralises 10 mL of M//3 Al_(2) (CO_(3))_(3) . Calculate the value of x .

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

If 100 mL of 1 N H_(2)SO_(4) is mixed with 100 mL of 1 M NaOH solution. The resulting solution will be