1 answer

You need to make 550ml of a 6.5% (w/v) sodium nitrite (NaNO2) solution as a food...

Question:

You need to make 550ml of a 6.5% (w/v) sodium nitrite (NaNO2) solution as a food preservative to prevent microbial growth. How many grams of NaNO2 (f.w.=68.995 g/mol) will you need? What is the molarity (M)of this solution?


Answers

A 6.5% (w/v) solution means 6.5 grams of sodium nitrite in 100 ml of solution.

So, using unitary method, if 100 ml of the solution contains 6.5 grams of sodium nitrite,

1 ml of solution will contain 6.5/100 grams of sodium nitrite

550 ml of solution will contain (6.5/100) x 550 grams of sodium nitrite.

this means 550 ml will contain 35.75 grams of sodium nitrite.

Molarity is defined as number of moles of solute per liter of solution.

The molecular weight of sodium nitrite is 68.995 grams. That means 1 mole of sodium nitrite weighs 68.995 grams.

number of moles = mass/ molecular mass

= 35.75 grams/  68.995 grams = 0.518 moles.

So, we know that 0.518 moles are present in 550 ml of the solution.

that means, 0.518/550 moles/ml. = 9.41 x 10-4 moles/ ml of solution.

1 litre = 1000 ml.

So 9.41 x 10-4 x 1000 moles/ litre of solution

= 0.94 moles/ litre.

Hence, the molarity is 0.94 M ( or moles/litre)

.

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