calculate the number of moles of potassium iodate (KIO) that were present in the 5.00mL aliquot of standar potassium iodate solution
Can someone please show me the workings to do questions like this one? thanks =]
You don't provide enough information; however, what you want is
moles = M x L = M x 0.005L = ?
To calculate the number of moles of potassium iodate present in a given solution, we need to know the molarity (concentration) of the solution.
Assuming we have the molarity (M) of the potassium iodate solution, we can use the following equation to calculate the number of moles:
Number of moles = Molarity (M) x Volume (L)
Given:
Volume = 5.00 mL = 5.00 x 10^-3 L (since 1 mL = 1 x 10^-3 L)
Let's say the molarity of the potassium iodate solution is 0.10 M. Using the equation mentioned above, we can calculate the number of moles as follows:
Number of moles = 0.10 M x 5.00 x 10^-3 L
= 5.00 x 10^-4 moles (or 0.0005 moles) of potassium iodate
So, in this case, there would be 0.0005 moles (or 5.00 x 10^-4 moles) of potassium iodate present in the 5.00 mL aliquot of the standard potassium iodate solution.
To calculate the number of moles of potassium iodate (KIO) present in the 5.00 mL aliquot of standard potassium iodate solution, you will need information about the concentration of the solution. Let's assume the concentration of the solution is given in units of moles per liter (M).
First, convert the volume of the aliquot from milliliters (mL) to liters (L). There are 1000 milliliters in a liter, so 5.00 mL is equal to 5.00/1000 = 0.005 L.
Next, you need to know the concentration of the potassium iodate solution. Let's assume it is given as 0.25 M.
Now, you can use the equation:
moles = concentration x volume
Plugging in the values, we have:
moles = 0.25 M x 0.005 L
moles = 0.00125 moles
Therefore, there were 0.00125 moles of potassium iodate (KIO) present in the 5.00 mL aliquot of the standard potassium iodate solution.