How much water must be added to 200 g of a 10% NaOH solution to obtain 8%
solution
you are diluting it 10/8 times=1.25 times.
That means, one part original solution, and .25 parts water.
that means then add 50 grams water.
check:
original salt mass=20g
new salt mass=.08*250=20g
To solve this problem, we can use the equation for dilution:
C1V1 = C2V2
where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume.
Let's plug in the given values:
C1 = 10% = 0.10
V1 = 200 g
C2 = 8% = 0.08
Now, we can solve for V2, the final volume of the solution:
(0.10)(200) = (0.08)V2
20 = 0.08V2
Divide both sides of the equation by 0.08:
V2 = 20 / 0.08
V2 = 250
So, the final volume of the solution required to obtain an 8% NaOH solution is 250 g.
Now, let's find how much water needs to be added. We can subtract the initial volume (200 g) from the final volume:
Water to be added = V2 - V1
Water to be added = 250 g - 200 g
Water to be added = 50 g
Therefore, 50 g of water must be added to 200 g of a 10% NaOH solution to obtain an 8% solution.
To solve this problem, we need to calculate the amount of water that needs to be added to the 200 g of 10% NaOH solution to obtain an 8% solution.
First, let's find out how much NaOH is present in the 200 g of 10% solution. The concentration of a solution is expressed as a percentage, which means that 10% of the total solution weight is NaOH.
NaOH (in grams) = 10% of total solution weight
NaOH = (10/100) * 200 g
NaOH = 20 g
Now, we can set up an equation to solve for the amount of water needed.
NaOH in the final solution = NaOH in original solution + NaOH added
The amount of NaOH in the final solution should be 8% of the total solution weight.
NaOH in the final solution = (8/100) * (200 g + amount of water added)
Substituting the values, we have:
20 g = (8/100) * (200 g + amount of water added)
To solve for the amount of water added, we can rearrange the equation:
amount of water added = (20 g / (8/100)) - 200 g
Simplifying further:
amount of water added = (20 g / 0.08) - 200 g
Now, let's calculate the value:
amount of water added = 250 g - 200 g
amount of water added = 50 g
Therefore, 50 grams of water must be added to the 200 g of 10% NaOH solution to obtain an 8% solution.