Answer:
1.7133 M
Explanation:
Given:
weight solute = 6.0 g NaBr
volume solution = 34.0 mL H¿O
***note: MW (molecular weight) of NaBr = 103 g/mol
Molarity = [tex]\frac{weight solute}{MW solute (L solution)}[/tex] = [tex]\frac{6.0 g}{103 (\frac{34.0ml}{1000})}[/tex] = 1.7133 mol/L or M