# How many grams of lithium sulfate (Li_2SO_4) are required to dissolve in 459 g of water to make a 1.23 m solution?

Jun 15, 2016

I assume you want $\text{molal}$ NOT $\text{molar}$ concentration.

#### Explanation:

We want a $1.23 \cdot m o l \cdot k {g}^{-} 1$ solution with respect to $L {i}_{2} S {O}_{4}$. Here, we deal with a molal concentration.

$\text{moles of lithium sulfate"/"kilograms of solvent}$ $=$ $1.23 \cdot m o l \cdot k {g}^{-} 1$

Thus $\text{moles of lithium sulfate}$

$=$ $0.459 \cdot k g \times 1.23 \cdot m o l \cdot k {g}^{-} 1$ $=$ $0.565 \cdot m o l$

We require a mass of $0.565 \cdot m o l \times 109.94 \cdot g \cdot m o {l}^{-} 1$ $=$ $62.12 \cdot g$.

This site reports the solubility of lithium sulfate in water. I have not gone thru it. Your given problem may be physically unrealistic.