Ejercicios FyQPortada del sitio > Chemistry exercises > Solutions > Molarity from % (m/V) concentration (8303)
Molarity from % (m/V) concentration (8303)
Domingo 8 de septiembre de 2024, por
Calculate the molarity of a sulfurous acid solution whose concentration is 
(m/V).
First, set a quantity of solution as the basis for calculation. Consider 100 mL of solution, which gives you 8 g of solute, 
, (these amounts are the “translation” of 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}}}](local/cache-vignettes/L384xH44/cebd2432beffb5b61436a43762fc2357-5a945.png?1732971541 "M_{\ce{H2SO3}} = 2\cdot 1 + 1\cdot 32 + 3\cdot 16 = \color[RGB]{0,112,192}{\bm{82\ \frac{g}{mol}}}"){kind=small}
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}}](local/cache-vignettes/L353xH51/2474f672e60f0538a05a977d56c8f8dc-0519f.png?1732971541 "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}}}}](local/cache-vignettes/L318xH48/30c4ea39edaee0d5ecc8e65e6338aed1-473c6.png?1732971541 "M = \frac{9.76\cdot 10^{-2}\ mol}{0.1\ L} = \fbox{\color[RGB]{192,0,0}{\bm{0.98\ \frac{mol}{L}}}}"){kind=small}