Mass of solute = 14,0 g
Molar mass ( NH4Br ) = 14 + 1 . 4 + 80 = 98 g/mol-1
volume of the solution = 150 mL = 0,15 L
Molarity of the solution = mol/L-1
[tex]M = \frac{m}{mm.V} [/tex]
[tex]M = \frac{14,0}{98.0,15} [/tex]
[tex]M = \frac{14,0}{14,7} [/tex]
[tex]M = 0,952 mol/L ^{-1} [/tex]
Hope this helps!
