Answer
The molarity of that $NaCl$ solution is equal to 0.250 M.
Work Step by Step
1. Determine the molar mass of this compound ($NaCl$), and setup the conversion factors:
Molar mass :
$Na: 22.99g $
$Cl: 35.45g$
22.99g + 35.45g = 58.44g
$ \frac{1 \space mole \space (NaCl)}{ 58.44 \space g \space (NaCl)}$ and $ \frac{ 58.44 \space g \space (NaCl)}{1 \space mole \space (NaCl)}$
$\frac{1000 \space mL}{1 \space L}$ and $\frac{1 \space L}{1000 \space mL}$
2. Calculate the number of moles $(NaCl)$
$ 5.85 \space g \times \frac{1 \space mole}{ 58.44 \space g} = 0.100 \space moles$
3. Convert the volume to liters:
$400 \space mL \times \frac{1 \space L}{1000 \space mL} = 0.400 \space L$
4. Find the concentration in mol/L (Molarity) $(NaCl)$:
$C(M) = \frac{n(moles)}{volume(L)}$
$ C(M) = \frac{ 0.100 \space moles}{ 0.400 \space L} $
$C(M) = 0.250 \space M$
