How much water must be added to 1 liter of a 5% saline to get a 2% saline solution
the amount of saline must be the same before and after mixing:
.05(1) + 0(x) = .02(1+x)
.02x = .03
x = 1.5
check:
5% of 1L = .05L saline
2% of 2.5L = .05L saline
or, you can say that since you want 2/5 the concentration, you need 5/2 the original volume, or 2.5L
To find out how much water must be added to a 1 liter of 5% saline to get a 2% saline solution, we can set up a proportion.
Let's first determine the amount of saline in the original 1 liter of 5% saline solution:
Saline in 1 liter of 5% saline = 1 liter * 5% = 0.05 liters.
To get a 2% saline solution, the saline content should be equal to 2% of the total volume of the solution. Let's assume the amount of water to be added is 'x' liters.
Saline in x liters of water = x liters * 2% = 0.02 liters.
Setting up the proportion:
0.05 liters / 1 liter = 0.02 liters / (1 liter + x liters).
Cross-multiplying:
0.05 liters * (1 liter + x liters) = 0.02 liters * 1 liter.
0.05 liters + 0.05x liters = 0.02 liters.
0.05x liters = 0.02 liters - 0.05 liters.
0.05x liters = 0.03 liters.
Dividing both sides by 0.05, we get:
x liters = 0.03 liters / 0.05.
x liters = 0.6 liters.
Therefore, 0.6 liters of water must be added to 1 liter of a 5% saline solution to obtain a 2% saline solution.
To determine the amount of water needed to dilute a 5% saline solution to a 2% saline solution, we'll need to use a formula based on the principle of mixtures:
C₁V₁ + C₂V₂ = C₃V₃
Where:
C₁ = concentration of the initial solution (5%)
V₁ = volume of the initial solution (1 liter)
C₂ = concentration of the water (0% saline concentration)
V₂ = volume of the water to be added (unknown)
C₃ = concentration of the final solution (2%)
V₃ = volume of the final solution (1 liter + V₂)
First, let's plug in the given values into the formula:
(5%)(1 liter) + (0%)(V₂) = (2%)(1 liter + V₂)
Now, let's simplify and solve for V₂:
0.05 liter + 0 = 0.02 liter + (0.02 liter)(V₂)
0 = 0.02 liter + (0.02 liter)(V₂) - 0.05 liter
0 = (0.02 liter)(V₂) - 0.03 liter
To isolate V₂, let's subtract 0.02 liter from both sides:
-0.02 liter = (0.02 liter)(V₂) - 0.03 liter - 0.02 liter
-0.02 liter = (0.02 liter)(V₂) - 0.05 liter
Now, we can divide both sides by 0.02 liter:
(-0.02 liter) / (0.02 liter) = (0.02 liter)(V₂) / (0.02 liter)
-1 = V₂
Therefore, you would need to add -1 liter of water to the 1 liter of 5% saline solution to obtain a 2% saline solution.
However, a negative volume doesn't make sense in this context. In practical terms, it means that you cannot dilute a 5% saline solution with water to achieve a 2% saline solution. To achieve a 2% saline solution, you would need to start with a lower concentration of saline or use alternative methods.