Question

1.)How many grams would you need of a sample known to be 99.81% AgN03 by mass?

You are asked to prepare 150.0 mL of 3.15×10−2 M AgNO3.

2.) An aqueous solution is 6.40% methanol CH3OH by mass, with d = 0.988 g/mL}

What is the molarity of CH3OHOH} in this solution?

Answer 1

#1 is answered above.
#2.
0.988 density = 0.988 g/mL.
So 1000 mL has a mass of 988 grams.
Only 6.4% of that is CH3OH so
988 x 0.0640 = 63 something is g CH3OH.
moles = 63/32 = about 2 M
You need to work through these estimates and use more accurate numbers.