What is the molarity of a solution of Mg(OH)2 if the solution contains 1.35 g in exactly 100 mL of solution???


A 15.0 mL solution of H2SO3 is neutralized by 12.0 mL of a standradized 1.00M NaOH solution. What is the concentration of the H2SO3 solution??????

What volume of a 1.25M stock solution of NaCl would be needed to make up 250 mL of a .25M NaCl solution

mols Mg(OH)2 = gams/molar mass

Then M = mols/L soln.

H2SO3 + 2NOH => Na2SO3 + 2H2O
mols NaOH = M x L = ?
mols H2SO3 = 1/2 mols H2SO3 (look at the coefficients in the balanced equation.)
M H2SO3 = mols H2SO3/L H2SO3.

Use c1v1 = c2v2
c = concn
v = volume
1.25M*v1 = 0.25*250
Solve for v1

mols Mg(OH)2 = grams/molar mass

1.35g/58.32
=.0231

m/L solution

.0231/100ml = .00231

????? is that right?

To find the molarity of a solution, we use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

Let's solve each problem step by step:

1. Molarity of Mg(OH)2 solution:
a. Convert the mass of Mg(OH)2 to moles.
Molar mass of Mg(OH)2 = 24.31 g/mol + (16.00 g/mol + 1.01 g/mol) * 2 = 58.33 g/mol
moles of Mg(OH)2 = mass / molar mass = 1.35 g / 58.33 g/mol ≈ 0.0231 mol

b. Convert the volume of the solution to liters.
volume of solution = 100 mL = 0.100 L

c. Calculate the molarity.
Molarity (M) = moles of solute / volume of solution = 0.0231 mol / 0.100 L = 0.231 M

Therefore, the molarity of the Mg(OH)2 solution is 0.231 M.

2. Concentration of H2SO3 solution:
a. Calculate the moles of NaOH used for neutralization.
moles of NaOH = concentration of NaOH * volume of NaOH = 1.00 M * 0.012 L = 0.012 mol

b. The stoichiometry of the neutralization reaction tells us that the number of moles of NaOH is equal to the number of moles of H2SO3. Therefore, the moles of H2SO3 in the solution is also 0.012 mol.

c. Convert the moles of H2SO3 to concentration.
volume of H2SO3 = 15.0 mL = 0.015 L
Concentration of H2SO3 = moles of solute / volume of solution = 0.012 mol / 0.015 L = 0.8 M

Therefore, the concentration of the H2SO3 solution is 0.8 M.

3. Volume of stock solution of NaCl needed:
a. Calculate the moles of NaCl needed for the desired concentration.
moles of NaCl = concentration of NaCl * volume of NaCl = 0.25 M * 0.250 L = 0.0625 mol

b. Calculate the volume of the stock solution needed.
concentration of stock NaCl solution = 1.25 M
Volume of stock solution needed = moles of NaCl needed / concentration of stock NaCl solution = 0.0625 mol / 1.25 M = 0.05 L = 50 mL

Therefore, you would need 50 mL of the 1.25 M stock solution of NaCl to make up 250 mL of a 0.25 M NaCl solution.

To calculate the molarity of a solution, you need to know the moles of solute and the volume of the solution.

For the first question:
1. Convert the given mass of Mg(OH)2 (1.35 g) to moles. Use the molar mass of Mg(OH)2 to convert grams to moles.
- The molar mass of Mg(OH)2 is 58.33 g/mol.
- Moles of Mg(OH)2 = mass (g) / molar mass (g/mol).

2. Calculate the volume of the solution in liters. Convert the given volume of the solution (100 mL) to liters. It's important to work with volume in liters to match the definition of molarity.
- Volume (L) = volume (mL) / 1000.

3. Compute the molarity (M) using the formula:
Molarity (M) = moles of solute / volume of solution (L).

For the second question:
1. Use the given information to find the moles of H2SO3 and NaOH. You know that the reaction between H2SO3 and NaOH is in a 1:1 ratio, meaning one mole of H2SO3 reacts with one mole of NaOH.
- Moles of H2SO3 = moles of NaOH.
- Moles of NaOH = concentration (M) x volume (L).

2. Calculate the molarity of the H2SO3 solution.
- Molarity of H2SO3 solution = moles of H2SO3 / volume of H2SO3 solution (L).

For the third question:
1. Use the given information to find the moles of NaCl needed in the final solution.
- Moles of NaCl = concentration (M) x volume (L).

2. Rearrange the equation to solve for the volume of the stock solution.
- Volume of stock solution (L) = moles of NaCl / concentration (M).

Keep in mind that it's important to use consistent units throughout your calculations (e.g., converting mL to L) to ensure accurate results.