Question

820 g Li2SO4 is dissolved into 2500 mL of solution. What is the molar concentration?

Answer 1

Molarity is defined as the moles of solute per liter of solution. $M=\frac{n}{V}$. Where M is molarity, n is the number of moles and V is the volume. First we must find the molar mass of $Li_2SO_4$ which is 109.98 g/mol

$Li_2SO_4= 2 \times Li_{Ar} + S_{Ar} + 4 \times O_{Ar}\\= (2 \times 6.941 + 32.1 + 4 \times 16.0) = 109.98 g/mol$

$820\ g\ Li_2SO_4 \times \frac{mol}{109.98\ g}= 7.46\ mol\ Li_2SO4$

Then we find the molarity using above equation

$M= \frac{n}{V} = \frac{7.46 mol}{2500ml} \times\frac{1000ml}{L} = 2.98\ M$