The volume of 4n hcl and 10 n hcl required to make 1 litre of 6n hcl is
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Answered by
76
Let x be the amount of 4N HCl you need. Therefore, 1-x is the amount of 10N HCl you need.
4x + 10(1-x) = 1(6)
4x + 10 - 10x = 6
- 6x = -4
6x = 4
x = 4/6 = 0.67 L
Therefore, 1-x = 1-0.67= 0.33 L
So, you need 0.67 L of 4N HCl and 0.33 L of 10N HCl to make 1 L of 6N HCl.
4x + 10(1-x) = 1(6)
4x + 10 - 10x = 6
- 6x = -4
6x = 4
x = 4/6 = 0.67 L
Therefore, 1-x = 1-0.67= 0.33 L
So, you need 0.67 L of 4N HCl and 0.33 L of 10N HCl to make 1 L of 6N HCl.
tiwaavi:
There is the mistake in the answer.
Answered by
52
Hello Dear.
Here is the answer---
→→→→→→→→→
Let the Volume of the 4 N HCl be x liter.
N₁ = 4 N
∴ Volume of 10 N HCl = Total Volume - Volume of 4 N HCl
= (1 - x) liter.
N₂ = 10
Volume of the HCl Produced(V) = 1 liter.
Normality(N) = 6 N.
We know,
N₁V₁ + N₂V₂ = NV
⇒ (4)(x) + (10)(1 - x) = (6)(1)
∴ 4x + 10 - 10x = 6
∴ 10 - 6x = 6
∴ 6x = 4
∴ x = 2/3
Thus, the Volume of the 4 N HCl is 2/3 liter.
∴ Volume of the 10 N HCl = 1 - x
= 1 - 2/3
= (3 - 2)/3
= 1/3 liter.
→→→→→→→→→
Hope it helps.
Have a Good Day.
Here is the answer---
→→→→→→→→→
Let the Volume of the 4 N HCl be x liter.
N₁ = 4 N
∴ Volume of 10 N HCl = Total Volume - Volume of 4 N HCl
= (1 - x) liter.
N₂ = 10
Volume of the HCl Produced(V) = 1 liter.
Normality(N) = 6 N.
We know,
N₁V₁ + N₂V₂ = NV
⇒ (4)(x) + (10)(1 - x) = (6)(1)
∴ 4x + 10 - 10x = 6
∴ 10 - 6x = 6
∴ 6x = 4
∴ x = 2/3
Thus, the Volume of the 4 N HCl is 2/3 liter.
∴ Volume of the 10 N HCl = 1 - x
= 1 - 2/3
= (3 - 2)/3
= 1/3 liter.
→→→→→→→→→
Hope it helps.
Have a Good Day.
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