I collected questions from my book which I don't understand plz help me I am stuck plzz.!!!!!!!!

Three different solutions containing the same solute at the same temperature are made to be saturated. Beaker 1 has 200.0 mL, beaker 2 has 500.0 mL and beaker 3 has 1.0 L of solution. Which of these has the highest concentration? Explain your answer.

Calculate the mass, in grams, of solute needed to make each solution 250.0 mL of a 0.100 mol/L CaCl2 solution.

How much of a 3.00 mol/L solution, in Litres, can you make from 675 g of glucose (C6H12O6)?

What is the concentration of sodium ions when 35.0 g of sodium phosphate, Na3PO4, is dissolved in 500. mL of solution? Include the dissociation equation for sodium phosphate.

What volume of a 5.00 mol/L stock solution is needed to make 2.00 L of a 0.125 mol/L solution?

Find the new concentration when 225 mL of a 3.00 mol/L solution has 575 mL of water added to it.

What volume of water must be added to 250.0 mL of a 0.650 mol/L calcium chloride solution to produce a 0.430 mol/L solution?

What is the new concentration if 35 mL of 0.15 mol/L iron (II) nitrate is mixed with 72 mL of a 0.60 mol/L of the same substance?

Three different solutions containing the same solute at the same temperature are made to be saturated. Beaker 1 has 200.0 mL, beaker 2 has 500.0 mL and beaker 3 has 1.0 L of solution. Which of these has the highest concentration? Explain your answer.

All three solution have the same concentration. Saturation means that the solvent has dissolved all of the solute that it can dissolve. Each beaker contains more solvent but it also contains more solute ( as much more as can be dissolved in the extra solvent) but the concentration in mols/L (OR grams/L) stays the same.

Calculate the mass, in grams, of solute needed to make each solution 250.0 mL of a 0.100 mol/L CaCl2 solution.

How many mols CaCl2 do you need? That's mols = M x L = 0.100 x 0.2500 = 0.02500 mols. Then grams CaCl2 = mols CaCl2 x molar mass CaCl2 = ?

How much of a 3.00 mol/L solution, in Litres, can you make from 675 g of glucose (C6H12O6)?
How many mols do you have in 675 g glucose. That's mols = g/molar mass = 675/180 = ? Then M = mols/L. You know M and mols, solve for L.

What is the concentration of sodium ions when 35.0 g of sodium phosphate, Na3PO4, is dissolved in 500. mL of solution? Include the dissociation equation for sodium phosphate.
Na3PO4 ==> 3Na^+ + [PO4]^3-
How many mols Na3PO4 do you have? mols = g/molar mass = 35.0/164 = about 4.7 estimated. (Check that 164--I may not remember it correctly). Then (Na3PO4) = mols/L = about 4.7 mols/0.500 L = ?
Then (Na^+) will be 3 times that since there are 3 Na^+ in 1 molecule of Na3PO4.

That's half of them. From what you have here, see what you can do with the others.

Sure, I'd be happy to help you with these questions! Let's go through each question one by one and explain how to solve them.

1. Three different solutions containing the same solute at the same temperature are made to be saturated. Beaker 1 has 200.0 mL, beaker 2 has 500.0 mL, and beaker 3 has 1.0 L of solution. Which of these has the highest concentration? Explain your answer.

To determine which beaker has the highest concentration, we need to look at the ratio of solute to solvent. The concentration of a solution is usually expressed in terms of moles of solute per liter of solution (mol/L) or molarity.

To compare the concentrations, we need to calculate the moles of solute in each beaker and then divide by the volume of the solution.

2. Calculate the mass, in grams, of solute needed to make each solution 250.0 mL of a 0.100 mol/L CaCl2 solution.

To calculate the mass of solute, we can use the formula:
Mass of solute = moles of solute x molar mass of solute.

First, we need to calculate the number of moles of CaCl2 using the given molarity (0.100 mol/L) and volume (250.0 mL). Then, we can multiply the number of moles by the molar mass of CaCl2 to obtain the mass in grams.

3. How much of a 3.00 mol/L solution, in liters, can you make from 675 g of glucose (C6H12O6)?

To determine the volume of the solution, we need to first calculate the number of moles of glucose using the given mass (675 g) and the molar mass of glucose (C6H12O6). Then, we can use the molarity (3.00 mol/L) to calculate the volume in liters.

4. What is the concentration of sodium ions when 35.0 g of sodium phosphate, Na3PO4, is dissolved in 500 mL of solution? Include the dissociation equation for sodium phosphate.

To determine the concentration of sodium ions, we need to consider the dissociation of sodium phosphate (Na3PO4) in water. Sodium phosphate dissociates into three sodium ions (Na+) and one phosphate ion (PO4^3-).

First, we calculate the moles of sodium phosphate using the given mass (35.0 g) and the molar mass of Na3PO4. Then, we can divide the moles of sodium ions by the volume of the solution to obtain the concentration in mol/L.

5. What volume of a 5.00 mol/L stock solution is needed to make 2.00 L of a 0.125 mol/L solution?

To prepare a solution with a desired concentration, we can use the formula:
Volume of stock solution = (Volume of desired solution x Desired concentration) / Stock concentration.

In this case, we want to prepare 2.00 L of a 0.125 mol/L solution using a stock solution with a concentration of 5.00 mol/L. Plugging these values into the formula will give us the required volume of the stock solution.

6. Find the new concentration when 225 mL of a 3.00 mol/L solution has 575 mL of water added to it.

To find the new concentration, we need to consider the dilution of the original solution. Dilution is a process of adding more solvent (water in this case) to the solution, which increases the solution's volume and lowers the concentration.

We can use the formula:
C1V1 = C2V2,
where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume.

By plugging in the given values, we can calculate the new concentration.

7. What volume of water must be added to 250.0 mL of a 0.650 mol/L calcium chloride solution to produce a 0.430 mol/L solution?

Similar to the previous question, this problem involves dilution. We need to calculate the volume of water necessary to achieve the desired concentration.

Using the formula C1V1 = C2V2, we can determine the volume of water needed to dilute the original solution to the desired concentration.

8. What is the new concentration if 35 mL of 0.15 mol/L iron (II) nitrate is mixed with 72 mL of a 0.60 mol/L of the same substance?

To find the new concentration after mixing two solutions, we need to consider the total volume and number of moles of each solution.

First, we calculate the moles of iron (II) nitrate (Fe(NO3)2) in each solution using the given concentration and volume. Then, we add the moles from both solutions and divide by the total volume to obtain the final concentration.