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2 years ago

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

2 answers:

7 0

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

Nataly_w [17]2 years ago

6 0

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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How many grams of solute are needed to order to prepare 100.00 mL of a 0.1000 M solution of a compound with a molecular weight o

Troyanec [42]

The grams of solute are required.

The mass of solute is 3.5 g

c = Molarity = 0.1 M

M = Molar mass = 350 g/mol

V = Volume of solution = 100 mL = 0.1 L

n = Number of moles

m = Mass of solute

Molarity is given by

$c=\dfrac{n}{V}\\\Rightarrow n=cV\\\Rightarrow n=0.1\times 0.1\\\Rightarrow n=0.01\ \text{moles}$

Molar mass is given by

$M=\dfrac{m}{n}\\\Rightarrow m=Mn\\\Rightarrow m=350\times 0.01\\\Rightarrow m=3.5\ \text{g}$

The mass of solute is 3.5 g

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2 years ago

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Answer:

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Explanation:

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