Answer
Hint: To solve this question we should know about:The molar concentration unit [ \[\dfrac{{mol}}{{{\text{ }}L}}{\text{, }}\left( M \right){\text{ }}\] ] is a widely used metric for concentration.It refers to the number of moles of the target substance (solute) dissolved in one liter of solution. Here's how to figure out the concentration.
Formula used:
Calculate the concentration:\[\dfrac{{\left( {Weight{\text{ }}of{\text{ }}1{\text{ }}liter{\text{ }}solution} \right) \times \left( {purity} \right)}}{{molecular{\text{ }}weight}}\]\[\dfrac{{\left[ {Specific{\text{ }}gravity{\text{ }}of{\text{ }}solution{\text{ }}\left( {\dfrac{g}{{mL}}} \right) \times 1,000{\text{ }}\left( {mL} \right) \times \dfrac{{Purity{\text{ }}\left( {\dfrac{w}{w}\% } \right){\text{ }}}}{{100}}} \right]}}{{Molecular{\text{ }}weight}}\]Complete answer:
As we know:The concentration acid you use in a lab is \[34\% \;\dfrac{w}{w}\;\] So a density of approx. \[1.17 \cdot g \cdot m{L^{ - 1}}\] We require \[125.g.HCl\; = \;\dfrac{{125.g}}{{36.46.g.mo{l^{ - 1}}}}\; = \;3.43.mol.\]For volume, we divide the molar quantity by the concentration:\[\dfrac{{3.43.mol}}{{10.9.mol.{L^{ - 1}}}} \times 1000.mL.{L^{ - 1}} = 315.mL\]The questioner, however, has not mentioned the concentration with which we are working. We're speculating.Working back, a volume of \[283.mL\;\] corresponds to \[12.2.mol.{L^{ - 1}}\] this is very concentrated muriatic acid. The concentration of the acid should have been specified as a boundary condition.Note:
In solutions containing up to 38% HCl, hydrochloric acid is generated (concentrated grade). Chemically, higher concentrations up to little over 40% are conceivable, but the evaporation rate is so high that extra measures, such as pressurisation and refrigeration, are required for storage and handling.