Question

What is the molarity of concentrated nitric acid that has $ \rho =1.41\cdot g\cdot m{{L}^{-1}} $ , and is 70% concentrated, m/m?

Answer 1

Hint: One of the most frequent measures used to quantify the concentration of a solution is molarity (M), which represents the number of moles of solute per litre of solution (moles/Liter). Molarity is a unit of measurement that may be used to determine the volume of a solvent or the quantity of a solute.

Complete answer:
In chemistry, the molarity (M) of a solution, which is the number of moles of solute per litre of solution, is commonly used to express its concentration. The molar concentration ( $ {{c}_{i}} $ ) is determined by multiplying the number of moles of solute ( $ {{n}_{i}} $ ) by the total volume (V) of the Solution
 $ c=\dfrac{n}{V}=\dfrac{N}{{{N}_{\text{A}}}V}=\dfrac{C}{{{N}_{\text{A}}}} $
The density of a material is its mass per unit volume (more specifically, its volumetric mass density; also known as specific mass). Although the Latin letter D can also be used, the sign for density is (the lowercase Greek letter $ \rho $ rho). Density is defined mathematically as mass divided by volume.
 $ \rho =\dfrac{m}{V} $
where the density is $ \rho $ , the mass is m, and the volume is V.
The given density of the given solution is $ 1.41 \mathrm{~g} / \mathrm{mL} $ .
 $ 1000 \mathrm{~g} $ of the given solution will have a volume $ =\dfrac{\text { Mass }}{\text { Density }}=\dfrac{1000}{1.41}=709 \mathrm{ml} $ .
The given mass percent of nitric acid is $ 70% $ .
 $ 1000 \mathrm{~g} $ of the given solution will have $ 1000\times \dfrac{70}{100}=700~\text{g} $ of nitric acid.
As a result The given molarity of solution is $ \dfrac{\text{ Mass of nitric acid }}{\text{ Molar mass of nitric acid }\times \text{ Volume of solution }}=\dfrac{700}{63\times 709}= $ $ 0.0156\text{M} $ .

Note:
The use of molar concentration in thermodynamics is frequently inconvenient since the volume of most solutions varies somewhat with temperature owing to thermal expansion. This difficulty is typically handled by utilising temperature adjustment factors or a temperature-independent concentration measure such as molality. The reciprocal amount denotes the dilution (volume) that may be found in Ostwald's dilution law.