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What is the weight (in g) of anhydrous Na2CO3 present in 250 mL of 0.25 N solution?

Answer 1

Olivia Anton

#"Concentration"="Moles of stuff"/"Volume of solvent"............#

And #"mass of stuff"="molar mass"xx"moles of stuff"#

Thus,...

#"mass"="molar mass"xx"moles of stuff"xx"volume"xx"concentration"#.

#=105.99*g*cancel(mol^-1)xx0.25*cancel(mol)xx0.250*cancelLxx0.25*cancel(mol)*cancel(L^-1)#

#="a little under 2 grams"..............#

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Answer 2

Noah Aldrich

To calculate the weight of anhydrous Na2CO3 present in 250 mL of 0.25 N solution, you can use the formula:

Weight (g) = Normality (N) × Equivalent weight (g/mol) × Volume (L)

The equivalent weight of Na2CO3 is 105.99 g/mol.

So, Weight (g) = 0.25 N × 105.99 g/mol × 0.250 L = 6.62475 g

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.