An 87.4 g sample of Li2SO4 (molar mass 109.94 g/mol) is dissolved in enough water to produce 1.15 L of solution.
Calculate the concentration of Li+ ions in solution.
Write your answers using three significant figures
To calculate the concentration of Li+ ions in the solution, we first need to determine the moles of Li2SO4 present in the sample.
Step 1: Calculate the moles of Li2SO4:
moles of Li2SO4 = mass of Li2SO4 / molar mass of Li2SO4
moles of Li2SO4 = 87.4 g / 109.94 g/mol
moles of Li2SO4 = 0.7948 mol
Step 2: Calculate the concentration of Li+ ions:
Since Li2SO4 dissociates into 2 moles of Li+ ions per mole of Li2SO4, the concentration of Li+ ions will be double the moles of Li2SO4.
concentration of Li+ ions = 2 * moles of Li2SO4 / volume of solution
concentration of Li+ ions = 2 * 0.7948 mol / 1.15 L
concentration of Li+ ions = 1.733 mol/L
Therefore, the concentration of Li+ ions in the solution is 1.733 mol/L.