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It is required to prepaire 100g of a 27.5% by mass solution of NaOH.calculate the nf of solution of NaOH.

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I am going to assume your term, "nf," which I do not recognize as an acceptable unit of the metric system, is meant to be molarity (M). So, it seems to me that you are asking for the molarity of the solution that is indicated in the problem.

To calculate the molarity of the solution of NaOH in this problem, one must apply the ratio, moles per liter of solution, the definition of molarity (moles/L).

A 27.5% solution means (27.5 grams) / (100 grams of solution), which in this case is equivalent to
27.5 g / 100 mL solution, because the density of water is 1.00 g/mL.

Since 100 g solution is specified, that means there are 27.5 g of NaOH in the solution (100 mL volume, or 0.100 L).

Moles of NaOH in solution = [27.5 g NaOH / 40.0 g mol NaOH/ mol NaOH] = 0.688 mol NaOH.

Therefore, molarity of NaOH = mol NaOH / L = (0.688 mol NaOH / 0.100 L) = 6.88 M NaOH.

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Posted on Jan 11, 2011

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