The density of methanol is .7918 g/ml/ How many grams of methanol do I have to add to 35.0ml of water to make a solution containing 25% methanol?
Is that mass percent?
You want
[x grams MeOH/(x g MeOH + 35g H2O)] = 0.25
Solve for x = grams methanol.
Then mass = volume x density. Substitute mass and density and solve for volume.
Post your work if you get stuck.
To calculate the amount of methanol needed to make a solution containing 25% methanol, you can use the equation:
% methanol = (grams of methanol / total grams of solution) * 100
Let's assume the total grams of the solution is equal to the sum of the grams of methanol and grams of water. We can calculate the grams of water by multiplying the volume (in milliliters) by the density of water, which is 1 g/ml.
Grams of water = volume of water (ml) * density of water (g/ml)
Grams of water = 35.0 ml * 1 g/ml
Grams of water = 35.0 g
Now let's calculate the grams of methanol required to make a 25% methanol solution.
% methanol = (grams of methanol / total grams of solution) * 100
25% = (grams of methanol / (grams of methanol + grams of water)) * 100
To find the grams of methanol, we can rearrange the equation:
grams of methanol = (25% / 100%) * (grams of methanol + grams of water)
Substituting the values we know:
grams of methanol = (0.25) * (grams of methanol + 35.0)
Now we can solve for grams of methanol:
grams of methanol = 0.25 * grams of methanol + 0.25 * 35.0
grams of methanol - 0.25 * grams of methanol = 0.25 * 35.0
0.75 * grams of methanol = 8.75
grams of methanol = 8.75 / 0.75
grams of methanol = 11.67 g
You will need to add approximately 11.67 grams of methanol to 35.0 ml of water to make a solution containing 25% methanol.
To answer this question, we can use the concept of calculating the mass of a solution using the density and volume of its components.
First, we need to determine the amount of water in the solution. Since the volume of water is given as 35.0 mL, we know that the mass of water is equal to its volume multiplied by its density. The density of water is approximately 1 g/mL, so the mass of water is:
Mass of water = Volume of water x Density of water
= 35.0 mL x 1 g/mL
= 35.0 g
Next, we need to calculate the amount of methanol required to make a solution containing 25% methanol. We can use the following equation:
% concentration = (mass of solute / (mass of solute + mass of solvent)) x 100
To find the mass of methanol, we need to rearrange the equation:
mass of solute = % concentration x (mass of solute + mass of solvent) / 100
In this case, the % concentration of methanol is 25%, and the mass of solvent is 35.0 g (which represents the mass of water in the solution). Let's substitute these values into the equation:
mass of methanol = 25% x (mass of methanol + 35.0 g) / 100
Now, we can solve the equation for the mass of methanol.
25/100 * (mass of methanol + 35.0 g) = mass of methanol
Simplifying the equation:
25/100 * mass of methanol + 25/100 * 35.0 g = mass of methanol
25/100 * mass of methanol - mass of methanol = -25/100 * 35.0 g
-75/100 * mass of methanol = -25/100 * 35.0 g
Dividing both sides of the equation by -75/100, we get:
mass of methanol = (-25/100 * 35.0 g) / (-75/100)
mass of methanol = (25/100 * 35.0 g) / (75/100)
mass of methanol = (0.25 * 35.0 g) / (0.75)
mass of methanol = 8.75 g / 0.75
mass of methanol ≈ 11.67 g
Therefore, you would need to add approximately 11.67 grams of methanol to 35.0 mL of water to make a solution containing 25% methanol.