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Relationship between molarity and normality (8380)
Lunes 27 de enero de 2025, por
Calculate the normality of the following solutions: a) 
(2 M) b) 
(0.4 M) c) 
(3 M) d) 
(1 M) e) 
(2 M)
Normality is defined as the number of equivalents of solute in one liter of solution. The mass of an equivalent is the molecular mass divided by the number of "H" or "OH" groups in the acid or base considered. This means that the relationship between normality and molarity is that normality is equal to molarity multiplied by the number of "H" or "OH" groups in the species.
a) ![\ce{NOH2}\ (2\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 4\ N}}](local/cache-vignettes/L219xH28/0738344972da2610cbfb98b6572fd9ed-342c1.png?1733007720 "\ce{NOH2}\ (2\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 4\ N}}"){kind=small}
b) ![\ce{KOH}\ (0.4\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 0.4\ N}}](local/cache-vignettes/L246xH28/231c6ba56c3a66f0d5cb08147fac806e-8fd43.png?1737657153 "\ce{KOH}\ (0.4\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 0.4\ N}}"){kind=small}
c) ![\ce{H_2SO_3}\ (3\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 6\ N}}](local/cache-vignettes/L224xH28/316e09d835185628eb3f2872f7045160-d9df6.png?1733007720 "\ce{H_2SO_3}\ (3\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 6\ N}}"){kind=small}
d) ![\ce{Al(OH)_3}\ (1\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 3\ N}}](local/cache-vignettes/L243xH28/b2dbaf95dd7ea26cea3bf959928e2faa-828c9.png?1733007720 "\ce{Al(OH)_3}\ (1\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 3\ N}}"){kind=small}
e) ![\ce{NaCl}\ (2\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 2\ N}}](local/cache-vignettes/L210xH28/c83eb18dd999260e0f60514be891e947-8023d.png?1737657153 "\ce{NaCl}\ (2\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 2\ N}}"){kind=small}
(because it is a salt with a 1:1 stoichiometry between its ions).