Given that the molar mass of NaNO3 is 85.00 g/mol, what mass of NaNO3 is needed to make 4.50 L of a 1.50 M NaNO3 solution? Use .

The molar mass of NaNO3 is 85.00 g/mol and if the 4.50 liter of 1.5 M (if this means moles) then the mass of NaNO3 should just be 85.0 x 1.5 = 127.5grams This appears to be the case since the concentration of the solution is already given at 1.5 moles if M = moles.

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Tringa0 Tringa0 · Answer 2 · 2019-12-13T11:02:43+01:00

Answer:

573.75 grams of sodium nitrate is needed to make 4.50 L of a 1.50 M sodium nitrate solution.

Explanation:

[tex]Moles (n)=Molarity(M)\times Volume (L)[/tex]

Moles of sodium nitrate= n

Volume of sodium nitrate solution = 4.5 L

Molarity of the sodium nitrate solution = 1.50 M

[tex]n=1.50 M\times 4.5 L=6.75 mol[/tex]

Mass = moles(n) × Molar mass

Molar mass of sodium nitrate = 85.00 g/mol

Mass of 6.75 moles of sodium nitrate = 6.75 mol × 85.00 g/mol = 573.75 g

573.75 grams of sodium nitrate is needed to make 4.50 L of a 1.50 M sodium nitrate solution.