Determine the molarity of sodium cations in solution when 6.21 g of sodium sulflate are mixed with 2.97 g of sodium nitrate in pure water to produce 350.0 mL of solution.......please explain every step. thanks!
moles Na2SO4 = grams/molar mass.
moles Na^+ is twice that.
moles NaNO3 = grams/molar mass.
moles Na^+ is same.
M Na^+ in soln = total moles Na^+/total volume in L.
6.21g/142.042 = 0.0437
2.97g/84.994 = 0.3494
M=moles/L
0.0437+0.3494/350.0= 1.12x10^-03
So I got = 1.12x10^-03
but my book says [Na+]= 0.350M
Where did I go wrong?
6.21g/142.042 = 0.0437
<If you re-read my instructions, you will note that Na^+ in Na2SO4 is TWICE moles Na2SO4; therefore, this should be 0.04372 x 2 = 0.08744
2.97g/84.994 = 0.3494
I suspect this is a typo. It should be 0.03494.
M=moles/L
0.0437+0.3494/350.0= 1.12x10^-03
(0.08744+0.03494/0.350 = 0.34966M which rounds to 0.350M to three s.f. Its a minor point but if you re-read my instructions, they were to use total volume in L (not mL).
So I got = 1.12x10^-03
but my book says [Na+]= 0.350M
Where did I go wrong?
1. You didn't double the Na2SO4.
2. You made a typo or an error in the second part (the NaNO3).
3. You didn't use volume in L.
Considering that you had specific instructions to work the problem, all of this can be laid to "not paying close attention to details." I hope this helps.
Thank you it has!
To determine the molarity of sodium cations in the solution, we first need to calculate the number of moles of sodium cations present.
Step 1: Calculate the number of moles of sodium sulfate
To convert the mass of sodium sulfate (Na2SO4) to moles, we need its molar mass. The atomic masses of sodium (Na), sulfur (S), and oxygen (O) are approximately 22.99 g/mol, 32.07 g/mol, and 16.00 g/mol, respectively.
Molar mass of Na2SO4 = (2 * 22.99 g/mol) + 32.07 g/mol + (4 * 16.00 g/mol) = 142.04 g/mol
Now, we can calculate the number of moles of sodium sulfate:
Moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4
= 6.21 g / 142.04 g/mol
Step 2: Calculate the number of moles of sodium nitrate
Using the same approach as above, we find the molar mass of sodium nitrate (NaNO3) to be approximately 85.00 g/mol.
Moles of NaNO3 = mass of NaNO3 / molar mass of NaNO3
= 2.97 g / 85.00 g/mol
Step 3: Calculate the total number of moles of sodium cations
Since both sodium sulfate and sodium nitrate contain sodium cations (Na+), we need to sum up the moles of sodium cations from both compounds.
Total moles of Na+ = moles of Na2SO4 + moles of NaNO3
Step 4: Calculate the molarity (M) of sodium cations
Molarity is defined as moles of solute per liter of solution. In this case, the solution volume is given as 350.0 mL. We can convert it to liters by dividing by 1000.
Molarity (Na+) = Total moles of Na+ / Solution volume in liters
= (Total moles of Na+) / (350.0 mL / 1000 mL/L)
Finally, you can substitute the calculated values for Total moles of Na+ into the formula to find the molarity of sodium cations in the solution.