https://ejercicios-fyq.com/ Ejercicios Resueltos, Situaciones de aprendizaje y VÍDEOS de Física y Química para Secundaria y Bachillerato es SPIP - www.spip.net https://ejercicios-fyq.com/local/cache-vignettes/L144xH25/siteon0-da713.png?1758361862 https://ejercicios-fyq.com/ 25 144 https://ejercicios-fyq.com/Molarity-from-m-V-concentration-8303 https://ejercicios-fyq.com/Molarity-from-m-V-concentration-8303 2024-09-08T03:42:27Z text/html es F_y_Q Concentration Molarity Percentage (mass/volume) Solutions SOLVED <p>Calculate the molarity of a sulfurous acid solution whose concentration is (m/V).</p> - <a href="https://ejercicios-fyq.com/Solutions-271" rel="directory">Solutions</a> / <a href="https://ejercicios-fyq.com/Concentration" rel="tag">Concentration</a>, <a href="https://ejercicios-fyq.com/Molarity" rel="tag">Molarity</a>, <a href="https://ejercicios-fyq.com/Percentage-mass-volume" rel="tag">Percentage (mass/volume)</a>, <a href="https://ejercicios-fyq.com/Solutions-660" rel="tag">Solutions</a>, <a href="https://ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>Calculate the molarity of a sulfurous acid solution whose concentration is <img src='https://ejercicios-fyq.com/local/cache-vignettes/L30xH19/86ee91cf864173a2378fbdeb1f3d916e-a72b9.png?1732971539' style='vertical-align:middle;' width='30' height='19' alt="8\ \%" title="8\ \%" /> (m/V).</math></p></div> <hr /> <div <div class='rss_ps'><p>First, set a quantity of solution as the basis for calculation. Consider 100 mL of solution, which gives you 8 g of solute, <img src='https://ejercicios-fyq.com/local/cache-TeX/ebddc00967ec960653ac0c88f6e17e8b.png' style="vertical-align:middle;" width="46" height="16" alt="\ce{H2SO3}" title="\ce{H2SO3}" />, (these amounts are the “translation” of <img src='https://ejercicios-fyq.com/local/cache-TeX/86ee91cf864173a2378fbdeb1f3d916e.png' style="vertical-align:middle;" width="30" height="19" alt="8\ \%" title="8\ \%" /> m/V). <br/> <br/> The moles corresponding to that mass of solute are obtained from the molecular mass of the acid: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/cebd2432beffb5b61436a43762fc2357.png' style="vertical-align:middle;" width="384" height="44" alt="M_{\ce{H2SO3}} = 2\cdot 1 + 1\cdot 32 + 3\cdot 16 = \color[RGB]{0,112,192}{\bm{82\ \frac{g}{mol}}}" title="M_{\ce{H2SO3}} = 2\cdot 1 + 1\cdot 32 + 3\cdot 16 = \color[RGB]{0,112,192}{\bm{82\ \frac{g}{mol}}}" /> <br/> <br/> Calculate the moles equivalent to the mass of solute: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/2474f672e60f0538a05a977d56c8f8dc.png' style="vertical-align:middle;" width="353" height="51" alt="8\ \cancel{g}\ \ce{H2SO3}\cdot \frac{1\ mol}{82\ \cancel{g}} = \color[RGB]{0,112,192}{\bm{9.76\cdot 10^{-2}\ mol}}" title="8\ \cancel{g}\ \ce{H2SO3}\cdot \frac{1\ mol}{82\ \cancel{g}} = \color[RGB]{0,112,192}{\bm{9.76\cdot 10^{-2}\ mol}}" /> <br/> <br/> 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): <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/30c4ea39edaee0d5ecc8e65e6338aed1.png' style="vertical-align:middle;" width="318" height="48" alt="M = \frac{9.76\cdot 10^{-2}\ mol}{0.1\ L} = \fbox{\color[RGB]{192,0,0}{\bm{0.98\ \frac{mol}{L}}}}" title="M = \frac{9.76\cdot 10^{-2}\ mol}{0.1\ L} = \fbox{\color[RGB]{192,0,0}{\bm{0.98\ \frac{mol}{L}}}}" /></p> </math></p></div>