Answer :
Answer:
The molarity is [tex]M = 0.7937 \ mol/L[/tex]
Explanation:
From the question we are told that
The mass of NaBr [tex]m_b = 12.25 \ g[/tex]
The volume of the solution is [tex]V = 150 m L = 150 *10^{-3 }L[/tex]
The number of moles of NaBr is
[tex]n_b = \frac{m_b}{M_b }[/tex]
Where [tex]M_b[/tex] is the molar mass of NaBr which is a constant with value [tex]M_b = 102.894 g/mol[/tex]
So
[tex]n_b = \frac{12.25}{102.894}[/tex]
[tex]n_b = 0.1191 \ mol[/tex]
The Molarity is mathematically evaluated as
[tex]M = \frac{n_b}{V}[/tex]
Substituting values
[tex]M = \frac{0.1191}{150 *10^{-3}}[/tex]
[tex]M = 0.7937 \ mol/L[/tex]