Asample of 61.8g of hyroboric acid(h3bo3), a weak acid, is dissolved in 1,000g of water to make a 1.0 molal solution. which would be the best procedure to determine the molarity if the solution?

Molarity = 1.0 M

Explanation:

Step 1:  Data given

Mass of H3BO3 = 61.8 grams

Mass of water = 1,000 grams = 1kg

Molality = 1.0 molal

Step 2: Calculate number of moles

1.0 molal = 1 mol of solute/ 1kg of solvent

Number of moles = molality * mass of the solvent = 1.0 mol/kg * 1kg = 1 mol

This means 61.8 grams of H3BO3 should be 1 mol

We can control this:

Number of moles = mass / molar mass

Number of moles H3BO3 = 61.8/ 61.83 g/mol = 1 mol

Step 3: Calculate the volume of water

Let's suppose the density of water is 1g/mL

Volume of water = mass / density

Volume = 1,000 grams / 1g/mL

Volume = 1,000 mL = 1.00 L

Step 4: Calculate molarity

Molarity = number of moles / volume

Molarity = 1 mol / 1.00 L

Molarity = 1.0 M

For dilute AQUEOUS solutions molality  ≅  molarity

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o