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Q: How many grams of LiCl are required to make 2.0 of 0.65?
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How many grams of LiCl are required to make 2.0 L of 0.65 M LiCl solution?

To calculate the number of grams of LiCl required, you need to use the formula: ( \text{grams} = \text{Molarity} \times \text{Volume} \times \text{Molar Mass} ) Given that the molar mass of LiCl is approximately 42.39 g/mol, plug in the values and calculate the grams of LiCl needed.


How many grams of LiCl would you use to make 20 ml of a 4M solution?

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.


A student weighed 0.550 grams of lithium chloride LiCl to use in a reaction How many moles is this?

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.


How many formula units of LiCl are equal to .648 moles?

42.394 grams.


How many gram gold require to make a gold ring?

2-3 grams is typically required.


How many toothpicks equal 40 grams?

The weight of a toothpick can vary, but on average, one toothpick weighs about 0.25 grams. Therefore, 160 toothpicks would be required to make up 40 grams.


How is a concentrated solution prepared?

3 Simple Steps!NOTE: Use this answer to make a solution if you need to dissolve a SOLID to make your solution. If instead you are making a solution by diluting another solution, see the Related Questions link for "How do you prepare a solution by diluting another solution?" You must know two things to make a solution: the number of moles of the solute, and the number of liters of solution. That is how the concentration "molarity" (M) is defined: the moles per liter:-- Molarity = moles of solute ÷ liters of solutionor-- Moles of solute = Molarity * liters of solutionNote: The volume is liters of solution, not solvent. That means that the total volume of the solution is what matters, not how much liquid you add.So, to make a solution of a specific concentration, we must figure out how many moles of solute we need. That's easy! Let me illustrate with a few simple examples. Then we'll find how many grams that corresponds to.--- Example 1: How many moles of lithium chloride (LiCl) are needed to make 1 liter of 0.5 M LiCl solution?--- Answer: In 1 liter of 0.5 M LiCl, there are 0.5 moles of LiCl. Therefore the answer is 0.5 moles LiCl.--- Example 2: How many more of sodium hydroxide (NaOH) are needed to make 2 liters of 0.5 M NaOH?--- Answer: In 1 liter of 0.5 M NaOH, there are 0.5 moles. So in 2 liters, there will be 1.0 moles. So the answer is 1.0 moles NaOH.--- Example 3: How many moles of silver nitrate (AgNO3) are needed to make 3.7 liters of 0.43 M AgNO3 solution?--- Answer: Same as before, but the math is more complicated. In 1 liter, there are 0.43 moles, so in 3.7 liters, there are 0.43 * 3.7 = 1.591 moles.Now we can't measure moles directly, so we need to find the number of grams of the solute so we can weigh it out. Once we've found the number of moles, we can convert that to grams from the molar mass of the substance (or atomic mass of the element). We need the periodic table for this (see the link below for a periodic table). Again, let me illustrate with a couple of examples. In fact I'll use the same examples as before.--- Example 1: How many grams of LiCl are needed to make 1 liter of 0.5 M LiCl?--- Answer: From before we figured out that we need 0.5 moles of LiCl. So how many grams is that? We first need to calculate the molar mass of LiCl. From the periodic table, the atomic weight of lithium (Li) is 6.941 grams per mole and the weight of chlorine (Cl) is 35.453 grams per mole. So the atomic mass of LiCl is: 6.941 + 35.453 = 42.394 grams per mole. So if we need 0.5 moles, than to find the grams, we just multiply: 0.5 moles * 42.394 grams/mole = 21.197 grams So, to make 1 liter of 0.5 M LiCl, we need to add 21.197 grams of HCl and add enough water to make the total volume equal 1 liter.--- Example 2: How many NaOH are need to make 2 liters of 0.5 M NaOH?--- Answer: From before, we know we need 1 mole of NaOH. The molar mass of NaOH is equal to: 22.9898 + 15.999 + 1.0079 = 39.9967 grams per mole So since we need 1 mole, we add 39.9967 grams of NaOH and add water until we have a total of 2 liters of solution.--- Example 3: How many grams of AgNO3 are needed to make 3.7 liters of 0.43 M acetic acid solution?--- Answer: Again, from before we know we need 1.591 moles of AgNO3. To find the number of grams, we must find the molar mass. So the molar mass of AgNO3 is: 107.868 + 14.007 + (3*15.999) = 169.873 g/mole So to find the number of grams, we just multiply: 1.591 moles * 169.873 g/mole = 270.268 grams. So we must add 270.268 grams of acetic acid, and then add enough water to make the total volume 3.7 liters. (Of course, that'd be really very expensive because silver nitrate is not cheap, and that a lot of it!)TO SUMMARIZE HOW TO MAKE A SOLUTION:1) From the concentration and the volume given to you in the problem, first find the number of moles of solute:Moles of solute = Concentration * Volume (or moles = molarity * liters)2) Then convert moles to grams using the molar mass:Number of grams = Number of moles * Molar mass (or grams = moles grams per mole)3) Then add water to the solid until the TOTAL volume of the solution is the correct amount you need.


How many loaned pairs of electrons does LiCl have?

LiCl does not have any loaned pairs of electrons. In LiCl, lithium donates one electron to chlorine to form an ionic bond, leading to a full outer shell for both elements.


How many moles are in 5L of 10 M LiCl?

There are 50 moles in 5 liters of 10 M LiCl (10 molar lithium chloride).


How many grams of protein is required by the body daily?

Seventeen


How many days are required for 200 grams of radon-222 to decay to 50.0 grams?

7.64 days


How many elements are there in LiCl?

There are two elements in LiCl: lithium (Li) and chlorine (Cl).