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.
Molarity = moles of solute/Liters of solution ( 250.0 ml = 0.250 liters ) Find moles. 61.7 grams LiCl (1 mole/42.391 grams) = 1.455 moles lithium chloride Molarity = 1.455 moles LiCl/0.250 liters = 5.82 M LiCl -------------------
To find the molarity, first calculate the number of moles of LiCl in 230 mL of water. Then, divide the moles of LiCl by the volume of water in liters (230 mL = 0.23 L) to get the molarity. In this case, 2.60 moles of LiCl in 0.23 L of water would result in a molarity of 11.30 mol/L.
To make a 4M solution in 20 ml, you would need 0.32 grams of LiCl. This can be calculated using the formula: moles = molarity x volume (in L), then converting moles to grams using the molar mass of LiCl.
The salt lithium chloride is LiCl. It's an Li+ ion and a Cl- ion.
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.
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.
To find the number of moles in 0.550 grams of LiCl, divide the mass by the molar mass of LiCl, which is approximately 42.4 g/mol. 0.550 g LiCl / 42.4 g/mol LiCl ≈ 0.013 mol LiCl. Therefore, the student has approximately 0.013 moles of LiCl.
Molarity = moles of solute/Liters of solution ( 250.0 ml = 0.250 liters ) Find moles. 61.7 grams LiCl (1 mole/42.391 grams) = 1.455 moles lithium chloride Molarity = 1.455 moles LiCl/0.250 liters = 5.82 M LiCl -------------------
You know because of solubility rules that LiCl disassociates 100% in water. So, knowing that molarity is equal to moles/liters Molarity LiCl = 1.97mol / 33.2 L Molarity = 0.059 M LiCl
To find the molarity, first calculate the number of moles of LiCl in 230 mL of water. Then, divide the moles of LiCl by the volume of water in liters (230 mL = 0.23 L) to get the molarity. In this case, 2.60 moles of LiCl in 0.23 L of water would result in a molarity of 11.30 mol/L.
To make a 4M solution in 20 ml, you would need 0.32 grams of LiCl. This can be calculated using the formula: moles = molarity x volume (in L), then converting moles to grams using the molar mass of LiCl.
There are 50 moles in 5 liters of 10 M LiCl (10 molar lithium chloride).
To calculate the number of moles of lithium chloride (LiCl) in 0.550 g, you need to divide the mass by the molar mass of LiCl. The molar mass of LiCl is approximately 42.39 g/mol (6.94 g/mol for Li and 35.45 g/mol for Cl). Therefore, 0.550 g of LiCl is approximately 0.013 moles.
C = 4.83 m LiClmole ratio = Xmoles = nA = soluteB = solventXA = nA___nA + nBsolution of LiCl in watersolute = LiClsolvent = water = H2Omolality = moles of solute = moles of solutekg of solvent 1.0 kg of solventAssume 1.000 kg H2O1.000 kg H2O * 1000 g * 1 mol H2O = 55.49 mol H2O1 kg 18.02 g H2OXA = nA___nA + nBSolve for nAXA( nA + nB ) = nAXAnA = XBnB = nAXAnB = nA - XAnAXAnB = nA( 1 - XA)XAnB___ = nA( 1 - XA)Plug in known amountsnA = 0.08 * 55.49 mol H2O = 4.83 mol LiCl1 - 0.08If molality = moles of solute1.0 kg of solventTHEN molality of LiCl = 4.83 mol LiCl = 4.83 m1.000 kg H2O