Question

2. Calculate the molarity of 500.0 grams of Li2S in a total volume of 4.44 liters?

Answer 1

Answer:

2.448 M ≅ 2.45 M.

Explanation:

Molarity (M) is the no. of moles of solute that dissolved in 1.0 L of the solution.

M = (no. of moles of solute)/(V of the solution (L).

M = (mass/molar mass) Li₂S / (V of the solution (L))

∴ M of Li₂S = (500.0 g/46.4 g/mol)/(4.44 L) = 2.448 M ≅ 2.45 M.

Answer 2

Answer:

The molarity of 500.0 grams of Li₂S in a total volume of 4.44 liters is 2.448 $\frac{moles}{liter}$

Explanation:

Molarity (M) is the number of moles of solute (substance that dissolves in a solvent with which it forms a solution, is in a smaller proportion) that are dissolved in a given volume.

Molarity is expressed by the following expression:

$M=\frac{number of moles of solute}{ volume of solution}$

Molarity is expressed in units $\frac{moles}{liter}$

Then, you must first know the amount of moles that represent the 500 g of Li₂S. For that, a rule of three is applied knowing the molar mass: if 46 g is 1 mol of Li₂S, 500 g of the compound, how many moles are they?

$moles=\frac{500 g*1mole}{46 g}$

moles≅10.87

It is now possible to calculate molarity by applying the definition:

$M=\frac{10.87 moles}{4.44 L}$

M= 2.448 $\frac{moles}{liter}$

The molarity of 500.0 grams of Li₂S in a total volume of 4.44 liters is 2.448 $\frac{moles}{liter}$