Question

commercial grade HCL solutions are typically 39.0% (by mass) HCL in water. Determine the molality and molarity of the HCL, if the solution has a density of 1.20 g/ml.

Answer 1

Concepts and reason

Molarity of the solution is defined as the amount of moles of a solute present in 1L of the solution.

The number of moles of solute present in 1kg of solvent is represented as molality.

Fundamentals

Molarity of the solution is calculated as follows:

${\rm{Molarity}}\,\,{\rm{ = }}\,\,\frac{{{\rm{moles}}\,\,{\rm{of}}\,{\rm{a}}\,{\rm{solute}}}}{{{\rm{Volume}}\,{\rm{in}}\,{\rm{litres}}}}$

Molality of a solution is calculated as follows:

${\rm{Molality}}\,\,{\rm{ = }}\,\,\frac{{{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{solute}}}}{{{\rm{kg}}\,\,{\rm{of}}\,{\rm{solvent}}}}$

$\begin{array}{l}\\{\rm{Assume that}}\,\,{\rm{volume}}\,\,{\rm{of}}\,{\rm{a solution}}\,\,{\rm{is}}\,\,{\rm{1}}{\rm{.00L}}\\\\{\rm{Mass}}\,\,{\rm{of}}\,\,{\rm{solution = }}\,\,{\rm{density}}\,\,{\rm{ \times }}\,{\rm{volume}}\\\\{\rm{Mass}}\,\,{\rm{of}}\,\,{\rm{solution = }}\,\,{\rm{1}}{\rm{.20}}\,\frac{{\rm{g}}}{{{\rm{mL}}}}\,{\rm{ \times }}\,\,\frac{{{\rm{1000}}\,{\rm{mL}}}}{{{\rm{1L}}}}\,\,{\rm{ \times }}\,\,{\rm{1}}{\rm{.00}}\,{\rm{L}}\\\\{\rm{Mass}}\,\,{\rm{of}}\,\,{\rm{solution = }}\,\,{\rm{1200}}\,{\rm{g}}\\\end{array}$

$\begin{array}{l}\\{\rm{Mass}}\,\,{\rm{of}}\,\,{\rm{HCl:}}\\\\{\rm{38\% }}\,\,{\rm{represents}}\,\,{\rm{that}}\,\,{\rm{100g}}\,\,{\rm{of}}\,{\rm{solution}}\,\,{\rm{contains}}\,{\rm{39g}}\,\,{\rm{of}}\,{\rm{Acid}}\,{\rm{present}}{\rm{.}}\\\\\,\,\,\,\,\,\,{\rm{Mass}}\,\,{\rm{of}}\,\,{\rm{HCl}}\,\,{\rm{ = }}\,\,\,{\rm{1200}}\,{\rm{g}}\,{\rm{solution}}\,\,{\rm{ \times }}\,\,\frac{{{\rm{39}}\,{\rm{g}}}}{{{\rm{100}}\,{\rm{g}}\,{\rm{solution}}}}\\\\\,\,\,\,\,\,\,{\rm{Mass}}\,\,{\rm{of}}\,\,{\rm{HCl}}\,\,{\rm{ = }}\,{\rm{468}}\,{\rm{g}}\,\\\end{array}$

$\begin{array}{l}\\{\rm{Moles}}\,\,{\rm{of}}\,\,{\rm{HCl:}}\\\\\,\,\,\,\,\,{\rm{Moles}}\,\,{\rm{of}}\,\,{\rm{HCl}}\,{\rm{ = }}\,\,\frac{{{\rm{mass}}}}{{{\rm{molecular}}\,{\rm{mass}}}}\\\\\,\,\,\,\,\,{\rm{Moles}}\,\,{\rm{of}}\,\,{\rm{HCl}}\,{\rm{ = }}\,\,\frac{{{\rm{468}}\,{\rm{g}}}}{{{\rm{36}}{\rm{.46}}\,{\rm{g}}}}\\\\\,\,\,\,\,\,{\rm{Moles}}\,\,{\rm{of}}\,\,{\rm{HCl}}\,{\rm{ = }}\,\,{\rm{12}}{\rm{.83}}\,{\rm{mol}}\,\\\end{array}$

$\begin{array}{l}\\{\rm{kg}}\,\,{\rm{of}}\,{\rm{solvent:}}\\\\\,\,\,\,\,{{\rm{m}}_{{\rm{solution}}}}{\rm{ = }}\,{{\rm{m}}_{{\rm{water}}}}{\rm{ + }}\,{{\rm{m}}_{{\rm{acid}}}}\\\\\,\,\,\,\,\,\,\,{{\rm{m}}_{{\rm{water}}}}\,{\rm{ = }}\,{{\rm{m}}_{{\rm{solution}}}}\,{\rm{ - }}\,{{\rm{m}}_{{\rm{acid}}}}\\\\\,\,\,\,\,\,\,\,{{\rm{m}}_{{\rm{water}}}}\,{\rm{ = }}\,{\rm{1200}}\,\,{\rm{ - }}\,\,{\rm{468}}\,\\\end{array}$

$\begin{array}{l}\\\,\,\,\,\,\,\,\,{{\rm{m}}_{{\rm{water}}}}\,{\rm{ = }}\,\,{\rm{0}}{\rm{.732}}\,{\rm{kg}}\,\left( {{\rm{732}}\,{\rm{g}}\,\,{\rm{ \times }}\,\,\frac{{{\rm{1}}\,{\rm{kg}}}}{{{\rm{1000g}}}}} \right)\\\\{\rm{Molality}}\,\,{\rm{ = }}\,\,\frac{{{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{solute}}}}{{{\rm{kg}}\,\,{\rm{of}}\,{\rm{solvent}}}}\\\\{\rm{Molality}}\,\,{\rm{ = }}\,\,\frac{{{\rm{12}}{\rm{.83}}\,{\rm{mol}}}}{{{\rm{0}}{\rm{.732}}\,{\rm{kg}}}}\\\\{\rm{Molality}}\,\,{\rm{ = }}\,{\rm{17}}{\rm{.53}}\,{\rm{m}}\\\end{array}$

$\begin{array}{l}\\\,{\rm{Molarity}}\,\,{\rm{of}}\,\,{\rm{HCl = }}\,\,\frac{{{\rm{Moles}}\,\,{\rm{of}}\,\,{\rm{solute}}}}{{{\rm{Volume}}\,\,{\rm{in}}\,\,{\rm{L}}}}\\\\\,\,{\rm{Moles}}\,\,{\rm{of}}\,\,{\rm{solute}}\,\,{\rm{ = }}\,\,{\rm{12}}{\rm{.83}}\,{\rm{mol}}\\\\\,\,\,\,\,\,\,\,{\rm{Volume}}\,\,{\rm{in}}\,\,{\rm{L}}\,\,{\rm{ = }}\,\,{\rm{1L}}\\\\{\rm{Molarity}}\,\,{\rm{of}}\,\,{\rm{HCl}}\,\,{\rm{ = }}\,\,\,\frac{{{\rm{12}}{\rm{.83}}\,{\rm{mol}}}}{{{\rm{1L}}}}\\\end{array}$

${\rm{Molarity}}\,\,{\rm{of}}\,\,{\rm{HCl}}\,\,{\rm{ = }}\,\,{\rm{12}}{\rm{.83}}\,{\rm{M}}$

Ans:

$\begin{array}{l}\\{\rm{Molality}}\,\,{\rm{ = }}\,\,{\rm{17}}{\rm{.53}}\,{\rm{m}}\\\\\,{\rm{Molarity}}\,{\rm{ = }}\,{\rm{12}}{\rm{.83}}\,{\rm{M}}\\\end{array}$

Answer 2

ManK Percent af HCL = 39.0 uhon Density of, the Rolutior)。1. 2.0 g/mL I.209 - 83.3 Lx - 0.0833 L 35 09 Molar Mas tl 36-69 36.44 9 0.0833レ Hence, the Molarit Molarity of tl i 12.8 M

Answer 3

given: commercial-grade HCl solutions are 39.0%(by mass) HCl in water.

the density of solution=1.20g/mL

volume of solution=1L=1000mL

molecular weight of HCl=36.5g/mol

molecular weight of H2O=18g/mol

mass of solution=density*volume=1.2*1000g=1200g

mass of HCl=0.39*1200=468g

no. of moles=468/36.5=12.82moles

molarity=no. of moles of solute/Vol of solution(in litres)

=12.82/1=12.82M

the correct answer is 12.8M