Question #95964

You need to start by using the definition of molarity to determine how many moles of $H C l$ can be found in $75 m L$ of solution.
${n}_{H C l} = C \cdot {V}_{s o l u t i o n} = 1.50 \frac{m o l e s}{L} \cdot 75 \cdot {10}^{- 3} L = 011 m o l e s$
Now, knowing that $H C l$'s molar mass is $36.5 \frac{g}{m o l e}$, we can determine the mass of $H C l$ dissolved in $75 m L$ by
${m}_{H C l} = {n}_{H C l} \cdot 36.5 \frac{g}{m o l e} = 0.11 m o l e s \cdot 36.5 \frac{g}{m o l e} = 4.0 g$