Answer
There is a total of 17.9 g of solute in 235 mL of that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 1.80 \space M \space LiCl$:
$1 \space L \space solution = 1.80 \space moles \space LiCl$
$\frac{1 \space L \space solution}{ 1.80 \space moles \space LiCl} $ and $\frac{ 1.80 \space moles \space LiCl}{1 \space L \space solution}$
2. Determine the molar mass of this compound (LiCl), and setup the conversion factors:
Molar mass :
$Li: 6.94g $
$Cl: 35.45g$
6.94g + 35.45g = 42.39g
$ \frac{1 \space mole \space (LiCl)}{ 42.39 \space g \space (LiCl)}$ and $ \frac{ 42.39 \space g \space (LiCl)}{1 \space mole \space (LiCl)}$
3. 1 L = 1000 mL, therefore:
$\frac{1 \space L}{1000 \space mL}$ and $\frac{1000 \space mL}{1 \space L}$
4. Use the conversion factors to calculate the mass of solute in $ 0.235$ L of that solution:
$ 235 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{ 1.80 \space moles \space LiCl}{1 \space L \space solution} \times \frac{ 42.39 \space g \space LiCl}{1 \space mole \space LiCl} = 17.9 \space g \space LiCl$
