Question

The relation between molarity (M) and molality (m) is given by:
(ρ = density of solution (mg/mL), M1= molecular weight of solute)
A) \[m = \dfrac{{1000M}}{{1000\rho - {M_1}}}\]
B) \[m = \dfrac{{1000\rho M}}{{1000\rho - M{M_1}}}\]
C) \[m = \dfrac{{1000\rho MM}}{{1000\rho - M{M_1}}}\]
D)\[m = \dfrac{{1000M}}{{1000\rho - M{M_1}}}\]

Answer 1

Hint: In chemistry, the concentration of the solution is very important. The solution strength is very important for medicinal chemistry and inhales solutions. There are a different way of representation in the concentration of solutions. There is mass by volume, volume percentage, molality, molarity, mole fraction, mass percentage, normality formality and parts per million.
Formula used:
The molarity of the solution value is equal to the ratio of the number of moles of the solute to the volume of the solution in litre. The symbol of molarity is M.
\[{\text{Molarity = }}\dfrac{{{\text{number of moles of the solute}}}}{{{\text{volume of the solution in litre}}}}\]
The molality of the solution depends on the number of moles of the solute and the weight of the solvent in litre. The molality of the solution is equal to the ratio of the number of moles of the solute to the weight of the solvent in litre. The symbol of molality is m.
\[{\text{Molality = }}\dfrac{{{\text{number of moles of the solute}}}}{{weight{\text{of the}}solvent{\text{in litre}}}}\]
Moles are defined as the given mass of the molecule and is divided by the molecular mass of the molecule.
\[{\text{moles}}\,{\text{ = }}\dfrac{{{\text{mass}}\,{\text{of}}\,{\text{the}}\,{\text{molecule}}}}{{{\text{molecular}}\,{\text{weight}}\,{\text{of}}\,{\text{the}}\,{\text{molecule}}}}\]
The molecular weight of the molecule depends on the atomic weight of the atom present in the molecule. The molecular weight of the molecule is equal to the sum of the molecular weight of the atom and the number of the respective atom in the molecule.
\[{\text{Molecular}}\,{\text{weight}}\,{\text{ = }}\,{\text{Number}}\,{\text{of}}\,{\text{the}}\,{\text{atoms}}\,{{ \times }}\,{\text{Atomic}}\,{\text{weight}}\,{\text{of}}\,{\text{the}}\,{\text{atom}}\]

Complete answer:
The given data is
ρ is the density of solution (mg/mL) and M1is molecular weight of solute
We consider ρ is the density of the solution.
The molecular weight of solute is M1
The molarity of the solution is M.
Because of the molarity consider,
The volume of the solution is \[1000{\text{ }}mL\].
\[The{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}solution{\text{ }} = {\text{ }}volume \times density\]
\[
   = 1000{\text{ }}mL \times \rho \;mg\;m{L^{ - 1}} \\
   = 1000\;\rho \;mg \\
 \]
The mass of the solution is \[1000\;\rho \;mg\].
The moles of the solute is M.
The mass of the solute is MM1.
The mass of the solvent is calculated,
\[ = 1000\;\rho \;mg - M{M_1}\]
We calculate the molality of the solution is given below,
\[{\text{Molality = }}\dfrac{{{\text{number of moles of the solute}}}}{{weight{\text{of the}}solvent{\text{in litre}}}}\]
\[
   = \dfrac{M}{{(1000\;\rho \; - M{M_1}) \times \dfrac{1}{{1000}}}} \\
   = \dfrac{{1000M}}{{(1000\;\rho \; - M{M_1})}} \\
 \]
According to the above discussion, we conclude the relation between molarity (M) and molality (m) (ρ is the density of the solution (mg/mL), M1 is the molecular weight of solute) is given by\[m = \dfrac{{1000M}}{{1000\rho - M{M_1}}}\]

Hence, the correct answer is option D.

Note:
The gram equivalent of the acid depends on the molecular mass of the acid divided by the basicity of the acid. The basicity of the acid is the number of hydrogen ions able to donate the acid in an aqueous medium. The gram equivalent of the acid is equal to the ratio of the molecular mass of the acid to the basicity of the acid. In acid base concept the hydrogen ion is very important.