Question

How many milliliters of a 1.09 M calcium acetate solution will be needed to react with 0.173 L of a 2.31 M solution of a sodium sulfate solution?

How many grams of iron(III) hydroxide could be produced from a reaction of 12.3 mL of 1.13 M sodium hydroxide with excess iron(III) chloride?

What is the molality of a solution made with 35.0 g of NaOH and 1250 g of water?

What is the change of boiling point for an aqueous solution made with 6.50 g of lithium chloride in 125 g of water?

What is the freezing point of a 0.125 m aqueous solution of ammonium phosphate?

Answer 1

i)

No. of moles of sodium sulphate = (Molarity)(Volume in L) = (2.31 M)(0.173 L) = 0.39963 mol

since, 1 mole of calcium acetate reacts with 1 mole of sodium sulphate. Therefore, No. of moles of calcium acetate reacted = No. of moles of sodium sulphate = 0.39963 mol

Volume of calcium acetate = (No. of moles of calcium acetate)/(Molarity)

= (0.39963 mol)/(1.09 M)

= 0.36663 L

= 366.63 mL (Ans)

ii)

Total no. of moles of sodium hydroxide (NaOH) = (Molarity)(Volume in L)

= (1.13 M)(12.3 x 10-3 L)

= 0.013899 mol

since, 1 mole of  iron(III) hydroxide (Fe(OH)3) requires 3 mole of sodium hydroxide.

Thus, total no. of moles of Fe(OH)3 formed = (0.013899 mol)/3 = 0.004633 mol

Mass of Fe(OH)3 = (No. of moles of Fe(OH)3)(molar mass of Fe(OH)3)

= (0.004633 mol)(106.867 g/mol)

= 0.495 g (Ans)

iii)

Mass of NaOH = 35 g

Molar mass of NaOH = 39.997 g/mol

No. of moles of NaOH (solute) = mass/molar mass

= (35 g)/(39.997 g/mol)

= 0.87506 mol

Mass of water (solvent) = 1250 g = 1250 x 10-3 Kg

Molality of NaOH solution = No. of mol of solute (NaOH)/ mass of solvent (water)

= (0.87506 mol)/(1250 x 10-3 Kg)

= 0.700 m

Molality of solution = 0.700 m (Ans)