# What is the molarity of NaCl when 3.46 g of NaCl is dissolved in water to form 50.0 mL of solution?

Jan 15, 2018

$\left[N a C l \left(a q\right)\right] = \cdot m o l \cdot {L}^{-} 1 \cong 1.20 \cdot m o l \cdot {L}^{-} 1$..
For $\text{molarity}$ we take the quotient...$\text{moles of solute"/"volume of solution}$...and we get an answer with units of $m o l \cdot {L}^{-} 1$..
And so....((3.46*g)/(58.44*g*mol^-1))/(50.0*mLxx10^-3*L*mL^-1)=??*mol*L^-1...
What are the values of $\left[N {a}^{+}\right]$ and $\left[C {l}^{-}\right]$?