127.17 g LiCl x 1 mol/42.4 g x 6.02x10^23 Form.Units/moles = 1.81x10^24 Formula Units.
127.17 g LiCl x 1 mol/42.4 g x 6.02x10^23 Form.Units/moles = 1.81x10^24 Formula Units.
The formula mass of the ionic compound lithium chloride, LiCl is 6.9 + 35.5 = 42.4.Amount of LiCl = 98.2/42.4 = 2.32molThere are 2.32 moles of formula unit in a 98.2g pure sample of LiCl.To get the numerical number, multiply the quantity in moles by the Avogadro's constant.
42.394 grams.
To determine the number of molecules in a sample of LiCl, we need to first calculate the number of moles using the molar mass of LiCl (42.39 g/mol). Next, we use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules. In this case, there are approximately (127.17 \text{ g} / 42.39 \text{ g/mol} \approx 3 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} ≈ 1.8 \times 10^{24}) molecules of LiCl in 127.17 g.
Formula: LiCl
To find the number of molecules of LiCl in a 127.17 g sample, you first need to convert the mass of LiCl to moles using its molar mass. Then, use Avogadro's number (6.022 x 10^23) to convert moles to molecules. Calculate the number of molecules of LiCl in the sample using these values.
LiCl
To calculate the number of molecules of LiCl in a 127.17g sample, you first need to determine the number of moles of LiCl in the sample using the molar mass of LiCl (6.94g/mol for Li and 35.45g/mol for Cl). Then, you can use Avogadro's number (6.022 x 10^23) to convert moles to molecules.
The salt lithium chloride is LiCl. It's an Li+ ion and a Cl- ion.
LiCl
LiCl
The compound formed between lithium and chlorine is lithium chloride, with the chemical formula LiCl.