4 Questions on molarity that I really don't understand. I have the formulas, but they don't seem to fit in to the question right. Please take a look, thanks!

1. Calculate the weight, in grams, of HNO3 present in 1.0 litre of 0.10M of solution.

2. What is the concentration, in moles per litre of 60.0g of NaOH in 500.0ml solution?

3. What volume of solution is required in each of the following if 4.0 moles of CuSO4 are used to make a 1.0M solution?

4. How can a solution be prepared to make 5.00L of a 6.00M solution of potassium hydroxide from a 8.00M stock solution?

1. You have 0.1 mol of HNO3 because you have one liter

H = 1
N = 14
O3 = 3*16 = 48
so
63 grams/mol
so
6.3 grams

The others are very similar.

what is the difference between moles(mol) and molarity(M)?

Sure! I can help you with these four questions on molarity. Let's go through each of them one by one.

1. Calculate the weight, in grams, of HNO3 present in 1.0 litre of 0.10M solution.
To solve this problem, we'll use the formula for calculating the weight of a solute in a given volume of solution:

Weight (g) = Molarity (M) x Molecular weight (g/mol) x Volume (L)

First, we need to find the molecular weight of HNO3. The molecular weight of HNO3 is calculated by adding up the atomic weights of its constituent atoms: H (1.01 g/mol) + N (14.01 g/mol) + O (16.00 g/mol) x 3 = 63.02 g/mol.

Now we can substitute the given values into the formula:
Weight of HNO3 = 0.10 M x 63.02 g/mol x 1.0 L = 6.302 grams

Therefore, the weight of HNO3 present in 1.0 litre of a 0.10M solution is 6.302 grams.

2. What is the concentration, in moles per litre, of 60.0g of NaOH in a 500.0ml solution?
To find the concentration in moles per litre, we'll use the formula:

Concentration (M) = (Mass of solute (g) / Molecular weight (g/mol)) / Volume of solution (L)

First, we need to find the molecular weight of NaOH. The molecular weight of NaOH is calculated by adding up the atomic weights of its constituent atoms: Na (22.99 g/mol) + O (16.00 g/mol) + H (1.01 g/mol) = 39.99 g/mol.

Now we can substitute the given values into the formula:
Concentration of NaOH = (60.0 g / 39.99 g/mol) / 0.500 L = 3.000 mol/L

Therefore, the concentration of NaOH in the 500.0ml solution is 3.000 moles per litre.

3. What volume of solution is required to make a 1.0M solution using 4.0 moles of CuSO4?
To find the volume of solution required, we'll rearrange the formula for molarity:

Volume of solution (L) = Moles of solute / Molarity (M)

Substituting the given values into the formula:
Volume of solution = 4.0 mol / 1.0 M = 4.0 litres

Therefore, to make a 1.0M solution using 4.0 moles of CuSO4, you would need a volume of 4.0 litres of solution.

4. How can a solution be prepared to make 5.00L of a 6.00M solution of potassium hydroxide from an 8.00M stock solution?
To prepare the desired solution, we can use the formula for dilution:

M1V1 = M2V2

where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Given that the initial molarity (M1) is 8.00M, the final molarity (M2) is 6.00M, and the final volume (V2) is 5.00L, we can rearrange the formula to solve for V1 (the initial volume):

V1 = (M2 x V2) / M1
V1 = (6.00 mol/L x 5.00 L) / 8.00 mol/L
V1 = 3.75 L

Therefore, to make a 5.00L 6.00M solution of potassium hydroxide from an 8.00M stock solution, you would need to measure 3.75L of the stock solution and dilute it with water up to a final volume of 5.00L.

I hope this helps you understand how to solve these types of molarity problems. Let me know if you have any further questions!