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What is the concentration of a solution made by combining 52 grams of NaCH3COO with 150 mL of water?

User Tophallen
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Answer: The concentration of a solution made by combining 52 grams of
NaCH_(3)COO^(-) with 150 mL of water is 4.22 M.

Step-by-step explanation:

Given: Mass = 52 grams

Volume = 150 mL (1 mL = 0.001 L) = 0.150 L

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

Hence, moles of
NaCH_(3)COO^(-) (molar mass = 82.034 g/mol) is as follows.


Moles = (mass)/(molar mass)\\= (52 g)/(82.034 g/mol)\\= 0.633 mol

Therefore, concentration of given solution is as follows.


Molarity = (moles)/(Volume(in L))\\= (0.633 mol)/(0.150 L)\\= 4.22 M

Thus, we can conclude that the concentration of a solution made by combining 52 grams of
NaCH_(3)COO^(-) with 150 mL of water is 4.22 M.

User Weirdpanda
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