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(a) Calculate the molarity of hydrogen chloride in a solution when $ 0.365{\text{ }}g $ of it has been dissolved in $ 100{\text{ }}mL $ of solution?
(b) $ 3.0{\text{ }}g $ of a salt of molecular weight $ 30 $ is dissolved in $ 250{\text{ }}ml $ water. The molality of solution is:

Answer
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Hint :Before going through the question let us first know about the key terms related to this question i.e. molarity and molality. The number of solvent moles present in $ 1{\text{ }}L $ of solution is defined as molarity. The volume of solutions comprises the denominator. The volume depends on the system temperature. The number of solute moles that are present in $ 1{\text{ }}kg $ of solvent is defined as molality. This unit is made up of mass conditions and is temperature independent.

Complete Step By Step Answer:
Now, let us solve the questions accordingly;
(a) The molarity of hydrogen chloride in a solution when $ 0.365{\text{ }}g $ of it has been dissolved in $ 100{\text{ }}mL $ of solution is:
Molarity $ \left( M \right) $ = $ \dfrac{n}{v} $
  $ M = $ Molar concentration
  $ n = $ Moles of solute
  $ v = $ Litres of solution
No. of moles $ (n) = $ $ \dfrac{{mass}}{{molar\,mass}} $
For $ HCl $ ;
Mass = $ 0.365{\text{ }}g $
Molar mass = $ {\text{36}}{\text{.5 g/mol }} $
Volume = $ 100mL = 0.1{\text{ }}L $
Thus, molarity = $ \dfrac{{0.365}}{{36.5 \times 0.1}} = 0.1M $
Therefore, the molarity of hydrogen chloride is $ 0.1M $ .
(b) $ 3.0{\text{ }}g $ of a salt of molecular weight $ 30 $ is dissolved in $ 250{\text{ }}ml $ water. The molality of solution is
The term for solution molality is:
  $ = \dfrac{{Number\,of\,moles\,of\,solute}}{{Weight\,of\,solvent\,in\,Kg}} \\
  = \dfrac{{Weight\,of\,solute}}{{Molar\,mass}} \times \dfrac{{1000}}{{Weight\,of\,solvent\,in\,Kg}} \\
 = \dfrac{3}{{30}} \times \dfrac{{1000}}{{250}} = 0.4M \\ $
Therefore, the molality of the solution is $ 0.4M $ .

Note :
Although molarity is usually used to express concentrations for solution reactions or for titrations, there is one drawback — molarity is the number of solute moles divided by solution volume, and the solution volume varies according to the density of the solution, depending on the temperature.