1.If 0.5l of 5M solution of CU(NO3) is diluted to a volume of 0.80l by adding water, what is the molarity of the resulting diluted solution?

2.How many ml of water will be required to dilute 11ml of 0.45Macid solution to a molarity of 0.1M?

3.Calculate the weight CuSO4.5H2O required for preparation of 1dm^3 of a solution whose concentration is 1000ppm of Cu.

1. To find the molarity of the resulting diluted solution, we can use the formula:

M1V1 = M2V2

where:
M1 = initial molarity of the solution (5M)
V1 = initial volume of the solution (0.5L)
M2 = molarity of the resulting diluted solution (unknown)
V2 = resulting volume of the diluted solution (0.80L)

Substituting the given values into the formula:

5M * 0.5L = M2 * 0.80L

Now, solve for M2:

M2 = (5M * 0.5L) / 0.80L
M2 = 3.125M

Therefore, the molarity of the resulting diluted solution is 3.125M.

2. To calculate the volume of water required to dilute a solution, we can use the formula:

M1V1 = M2V2

where:
M1 = initial molarity of the solution (0.45M)
V1 = initial volume of the solution (11mL)
M2 = desired molarity of the diluted solution (0.1M)
V2 = resulting volume of the diluted solution (unknown)

We need to find V2, the volume of the diluted solution. Therefore, we rearrange the formula to solve for V2:

V2 = (M1 * V1) / M2

Substituting the given values into the formula:

V2 = (0.45M * 11mL) / 0.1M
V2 = 4.95mL

Therefore, 4.95mL of water will be required to dilute 11mL of 0.45M acid solution to a molarity of 0.1M.

3. To calculate the weight of CuSO4.5H2O required for a solution with a concentration of 1000ppm of Cu (parts per million), we need to consider the molar mass of CuSO4.5H2O and convert ppm to grams.

1 ppm = 1 mg/L

To convert from ppm to grams:

1 ppm = 1 mg/L * 1 g / 1000 mg
1 ppm = 0.001 g/L

Now, we need to find the concentration of Cu in grams per liter (g/L) since the unit for parts per million is mg/L.

Concentration of Cu = 1000 ppm * 0.001 g/L
Concentration of Cu = 1 g/L

Since the molar mass of CuSO4.5H2O is 249.7 g/mol, we can calculate the weight of CuSO4.5H2O required for 1 dm^3 (1000 mL) of the solution:

Weight of CuSO4.5H2O = Concentration of Cu * Volume of Solution
Weight of CuSO4.5H2O = 1 g/L * 1000 mL
Weight of CuSO4.5H2O = 1000 g

Therefore, 1000 grams of CuSO4.5H2O is required for the preparation of 1 dm^3 (1000 mL) of a solution with a concentration of 1000 ppm of Cu.