We’re being asked to calculate the mass of NaCl needed to **prepare 0.250 L of 0.800 M NaCl **solution. To do this, we will use the formula for molarity. Recall that *molarity* is the ratio of the moles of solute and the volume of solution (in liters). In other words:

$\overline{){\mathbf{Molarity}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{=}}\frac{\mathbf{}\mathbf{mole}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Liters}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

We first need to determine the number of moles of NaCl from molarity and volume

Given: Molarity (^{mol}/_{L}) = 0.800 ^{mol}/_{L}

Volume = 0.250 L

Let's say we want to make 0.250 L of an aqueous solution with [NaCl] = 0.800 M. What mass of the solute, NaCl, would we need to make this solution?