How many sodium ions are present in 325 mL of 0.850 M Na2SO4?
number ions=.325*2*.85 moles
that is in moles, to get the absolute number, multiply by avagradro's number
To determine the number of sodium ions present in the given solution, you need to use the molarity (M) and volume (in liters) of the solution.
Step 1: Convert the volume from milliliters (mL) to liters (L):
325 mL = 325/1000 L = 0.325 L
Step 2: Calculate the number of moles of Na2SO4 using the formula:
moles = molarity x volume
moles of Na2SO4 = 0.850 M x 0.325 L = 0.27625 moles
Step 3: For every 1 mole of Na2SO4, there are 2 moles of sodium ions (Na+).
Number of moles of Na+ ions = 2 x 0.27625 moles = 0.5525 moles
Step 4: Convert moles to number of ions using Avogadro's number (6.022 x 10^23 ions/mole):
Number of Na+ ions = 0.5525 moles x (6.022 x 10^23 ions/mole) = 3.325 x 10^23 Na+ ions
Therefore, there are approximately 3.325 x 10^23 sodium ions in 325 mL of 0.850 M Na2SO4 solution.
To determine the number of sodium ions present in a solution, you need to know the concentration and volume of the solution, as well as the chemical formula of the compound.
In this case, we have 325 mL of a 0.850 M Na2SO4 solution. To calculate the number of sodium ions, we first need to find the number of moles of Na2SO4 using the equation:
moles = concentration (M) x volume (L)
Given that the volume is 325 mL, we need to convert it to liters by dividing it by 1000:
volume (L) = 325 mL / 1000 = 0.325 L
Now we can calculate the number of moles:
moles = 0.850 M x 0.325 L = 0.27625 moles
Since each formula unit of Na2SO4 contains 2 sodium ions, we can multiply the number of moles by 2 to determine the number of moles of sodium ions:
moles of sodium ions = 0.27625 moles x 2 = 0.5525 moles
Finally, we can convert moles to the number of sodium ions using Avogadro's number (6.022 x 10^23 ions/mol):
number of sodium ions = moles of sodium ions x Avogadro's number
number of sodium ions = 0.5525 moles x 6.022 x 10^23 ions/mol
Thus, there are approximately 3.327 x 10^23 sodium ions in 325 mL of 0.850 M Na2SO4 solution.