What volume of a solution that is 15.7 % (w/w) lolinic acid (density = 1.64 g/mL) contains 4.25 g of lolinic acid15.
What is the molarity of a 20.2 % (w/w) solution of rossonic acid (423.6 g mol-1) that has a density of 1.78 g mL-1?
To determine the volume of a solution that contains a specific amount of a solute, we can use the formula:
Volume = Mass of solute / (Density of solution x Concentration of solute)
For the first question, we are given the mass of lolinic acid (4.25 g), the concentration of lolinic acid (15.7% w/w), and the density of the solution (1.64 g/mL). Let's substitute these values into the formula:
Volume = 4.25 g / (1.64 g/mL x 15.7%)
But before calculating, we need to convert the concentration from percentage to a decimal. To do this, divide the percentage by 100:
Concentration of solute = 15.7% / 100 = 0.157
Now, we can calculate the volume:
Volume = 4.25 g / (1.64 g/mL x 0.157) = 16.54 mL
Therefore, the volume of the solution containing 4.25 g of lolinic acid is 16.54 mL.
For the second question, we are given the concentration of rossonic acid (20.2% w/w), the molar mass of rossonic acid (423.6 g/mol), and the density of the solution (1.78 g/mL). To find the molarity of the solution, we need to calculate the moles of rossonic acid first.
Moles of Rossonic Acid = Mass of Rossonic Acid / Molar Mass of Rossonic Acid
Given that the mass of the solution is not mentioned, we can assume that we have 100 g of the solution. Therefore, the mass of the rossonic acid can be calculated as follows:
Mass of Rossonic Acid = 20.2% x Mass of Solution = 20.2 g (since we assumed 100 g)
Now, let's substitute these values into the formula:
Moles of Rossonic Acid = 20.2 g / 423.6 g/mol = 0.0477 mol
Next, we can calculate the volume of the solution using the formula mentioned earlier:
Volume = Mass of Solution / Density of Solution
Volume = 100 g / 1.78 g/mL = 56.18 mL
Finally, we can calculate the molarity of the solution:
Molarity = Moles of Rossonic Acid / Volume of Solution (in liters)
Convert the volume to liters by dividing it by 1000:
Molarity = 0.0477 mol / (56.18 mL / 1000) = 0.850 M
Therefore, the molarity of the 20.2% w/w solution of rossonic acid is 0.850 M.