what is the total mass of LiBr required to make 500.0 grams of an aqueous solution of LiBr that has a concentration of 338 ppm?
388 = x grams/500 x 1,000,000
I too come from castlelearning XDD
see other post.
Well, you know what they say - when in doubt, add some humor! So, let's dive into your question.
To calculate the mass of LiBr required, we need to consider its concentration in parts per million (ppm). But first, let me ask you this: Why did the chemist like nitrates so much? Because they're cheaper than day rates! Okay, now back to your question.
The concentration of 338 ppm means that for every 1 million parts of the solution, 338 parts are LiBr. So, if we have 500.0 grams of solution, we know that:
1,000,000 parts = 500.0 grams
338 parts = x (the mass of LiBr)
To find the mass of LiBr, we can set up a proportion:
(338 parts / 1,000,000 parts) = (x grams / 500.0 grams)
Solving for x, we get:
x = (338/1,000,000) * 500.0
x ≈ 0.169 grams
Therefore, the total mass of LiBr required to make 500.0 grams of the solution is approximately 0.169 grams. That's a tiny amount, just like the chemical humor I sprinkle in!
To find the total mass of LiBr required, we need to use the concentration of the solution and the mass of the solution.
First, let's review the concept of concentration. Concentration is typically expressed in terms of mass/volume (e.g., grams per liter) or moles/volume (e.g., moles per liter). In this case, the concentration is given in parts per million (ppm), which means the mass of the solute per million parts of the total solution.
To calculate the total mass of LiBr required, we can use the equation:
Mass of solute = Concentration × Volume of Solution
In this case, we are given the concentration in ppm and the mass of the solution. However, the volume of the solution is not provided. To solve this, we need additional information.
If we assume the solution is made by dissolving 500.0 grams of LiBr in water to make a 500.0 gram solution (where the mass of the solvent, water, is negligible), then the total mass of the solution is 500.0 grams.
Now, let's calculate the mass of LiBr required using the provided concentration:
Mass of LiBr = Concentration × Volume of Solution
Since ppm represents the number of parts of solute per million parts of solution, we can convert ppm to a fraction by dividing by one million:
Concentration (in fraction) = Concentration (in ppm) / 1000000
Substituting the given concentration in ppm (338 ppm) into the equation, we get:
Concentration (in fraction) = 338 ppm / 1000000 = 0.000338
Since the volume of the solution is not provided, we'll assume it to be 1 liter (1000 milliliters) to simplify the calculation.
Mass of LiBr = Concentration × Volume of Solution
= 0.000338 × 1000 grams
= 0.338 grams
Therefore, the total mass of LiBr required to make 500.0 grams of an aqueous solution with a concentration of 338 ppm is 0.338 grams.