Molatity: 1) An aqueous solution of sodium hydroxide has a concentration of 0.26

Question

Molatity:

1) An aqueous solution of sodium hydroxide has a concentration of 0.263 molal.

The percent by mass of sodium hydroxide in the solution is  %.

2) An aqueous solution of magnesium sulfate has a concentration of 0.432 molal.

The percent by mass of magnesium sulfate in the solution is  %.

3) An aqueous solution is 13.0% by mass ethanol, CH3CH2OH, and has a density of 0.978 g/mL.

The molality of ethanol in the solution is  m.

Mole fraction/mass percent:

1) An aqueous solution is 26.0 % by mass ammonia, NH3, and has a density of 0.904 g/mL.

The mole fraction of ammonia in the solution is  .

2)An aqueous solution is 36.0 % by mass silver nitrate, AgNO3, and has a density of 1.41 g/mL.

The mole fraction of silver nitrate in the solution is  .

3) The mole fraction of iron(III) acetate, Fe(CH3COO)3, in an aqueous solution is 2.04×10-2 .

The percent by mass of iron(III) acetate in the solution is  %.

Explanation / Answer

Answer – Molarity-

1)Given, [NaOH] = 0.263 M , means 0.260 moles of NaOH in the 1000 mL or 1 L solution

We know water has density = 1.00 g/mL

So, 1000 mL = 1000 g of water

Mass of NaOH = 0.263 moles * 40.0 g/mol

                      = 10.52 g

Percent by mass of NaOH = mass of NaOH / mass of solution*100 %

                                            = 10.52 g / 1000 g * 100 %

                                            = 1.052 %

2) [MgSO4] = 0.432 M , means 0.432 moles of MgSO4 in the 1000 mL or 1 L solution

We know water has density = 1.00 g/mL

So, 1000 mL = 1000 g of water

Mass of MgSO4 = 0.432 moles * 120.37 g/mol

                            = 51.98 g

Percent by mass of MgSO4= mass of MgSO4/ mass of solution*100 %

                                            = 51.98 g / 1000 g * 100 %

                                            = 5.2 %

3) Percent by mass of CH3CH2OH = 13.0 % , density = 0.978 g/mL

Assume 100 mL of solution,

So mass of solution = 100 mL * 0.978 g/mL

                                 = 97.8 g

So, mass of CH3CH2OH –

Percent by mass of CH3CH2OH = mass of CH3CH2OH / mass of solution*100 %

So, mass of CH3CH2OH = Percent by mass of CH3CH2OH * mass of solution / 100 %

                                        = 13.0 % * 97.8 g / 100 %

                                        = 12.7 g

Moles of CH3CH2OH = 12.7 g / 46.068 g.mol-1

                                    = 0.276 moles

So mass of water = 97.8-12.7 = 85.1 g

So mass of water in kg = 0.0851 kg

We know molality = moles of solute / kg of solvent

                               = 0.276 moles / 0.0851 kg

                                = 3.24 m

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