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

i'm assuming that the correct answer is indeed the first one/ "a"..

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.