Molarity from % (m/V) concentration (8303)

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Calculate the molarity of a sulfurous acid solution whose concentration is 8\ \% (m/V).

P.-S.

First, set a quantity of solution as the basis for calculation. Consider 100 mL of solution, which gives you 8 g of solute, \ce{H2SO3}, (these amounts are the “translation” of 8\ \% m/V).

The moles corresponding to that mass of solute are obtained from the molecular mass of the acid:

M_{\ce{H2SO3}} = 2\cdot 1 + 1\cdot 32 + 3\cdot 16 = \color[RGB]{0,112,192}{\bm{82\ \frac{g}{mol}}}

Calculate the moles equivalent to the mass of solute:

8\ \cancel{g}\ \ce{H2SO3}\cdot \frac{1\ mol}{82\ \cancel{g}} = \color[RGB]{0,112,192}{\bm{9.76\cdot 10^{-2}\ mol}}

The molarity is the ratio between the moles of solute and the volume of the solution, expressed in liters, i.e., 0.1 L (because you considered 100 mL at the beginning):

M = \frac{9.76\cdot 10^{-2}\ mol}{0.1\ L} = \fbox{\color[RGB]{192,0,0}{\bm{0.98\ \frac{mol}{L}}}}

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