This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.
1 mole is equal to 6.022x10²³ particles as given, so:

<h3>
Answer:</h3>
2.49 mol
Let me know if you have any questions.
The average weight of an atom of an element, formerly based on the
weight of one hydrogen atom taken as a unit or on 1/16 (0.0625) the
weight of an oxygen atom, but after 1961 based on 1/12 the weight of the
carbon-12 atom.
Jupiter------------------------
I think I have done a question like this before and i'm pretty sure your answer would be the 2nd one 1s22s22p4. I'm not 100% sure but try it at least.
Answer:
We need 4.28 grams of sodium formate
Explanation:
<u>Step 1:</u> Data given
MW of sodium formate = 68.01 g/mol
Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L
pH = 3.74
Ka = 0.00018
<u>Step 2:</u> Calculate [base)
3.74 = -log(0.00018) + log [base]/[acid]
0 = log [base]/[acid]
0 = log [base] / 0.42
10^0 = 1 = [base]/0.42 M
[base] = 0.42 M
<u>Step 3:</u> Calculate moles of sodium formate:
Moles sodium formate = molarity * volume
Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles
<u>Step 4:</u> Calculate mass of sodium formate:
Mass sodium formate = moles sodium formate * Molar mass sodium formate
Mass sodium formate = 0.063 mol * 68.01 g/mol
Mass sodium formate = 4.28 grams
We need 4.28 grams of sodium formate