What mass of sodium oxalate na2c2o4 is needed to prepare 0.250 l of a 0.100 m solution?

molarity = number of moles of solute / liters of solution
number of moles = molarity * liters of solution
number of moles of Na2C2O4 = 0.1 * 0.25 = 0.025 moles

Now, from the periodic table:
mass of Na = 22.9 grams
mass of C = 12 grams
mass of O = 16 grams
molar mass of Na2C2O4 = 2(22.9) + 2(12) + 4(16) = 133.8 grams
Therefore, one mole is equal to 133.8 grams. To know the mass of 0.025 moles, all you have to do is cross multiplication as follows:
mass = (0.025*133.8) / 1 = 3.345 grams

arihsiddiqui04 arihsiddiqui04 · Answer 2 · 2020-03-22T23:11:01+01:00

arihsiddiqui04 arihsiddiqui04

22-03-2020

Answer:

3.35 g Na2C2O4

Explanation:

They gave us the molarity and the liters of solution so using the relationship (M=mol/V) we can get the moles of solution.

0.250 L (0.100 mol/L) = 0.0250 mol Na2C2O4

Now that we have the moles of sodium oxalate, we can use its molar mass (134 g/mol) to convert to grams.

0.0250 mol (134.0 g/mol) -=3.35 grams (rounded to three sig figs

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