The mole fraction of an aqueous solution of sodium bromate is 0.249. Calculate t

Question

The mole fraction of an aqueous solution of sodium bromate is 0.249. Calculate the molarity (in mol/L) of the sodium bromate solution, if the density of the solution is 1.05 g mL^-1. Determine the mole fraction of sodium periodate in a 3.40 M aqueous solution of sodium periodate. The density of the solution is 1.09 g mL^-1. Determine the mass (in g) of Co(OH)_2 that is produced when 317 mL of a 5.99 times 10^-2 M CoCl_2 solution completely reacts with 864 mL of a 1.29 times 10^-2 M NaOH solution according to the following balanced chemical equation. CoCl_2(aq) + 2NaOH(aq) - Co(OH)_2(s) + 2NaCl(aq)

Explanation / Answer

1. a. The sodium bromate solution comprises:

0.249 fraction of NaBrO3 and 0.751 fraction of water. If we consider 100 mL of the solution:

100 mL solution x 1.05 g/1 mL = 105 g of solution, from which 24.9% is NaBrO:

105 g x 24.9% = 26.145 g NaBrO3 x 1 mol NaBrO3/151 g = 0.173 moles of the compound.

Therefore: 0.173 moles/0.1 L = 1.73 M

2. Let's consider 1 L of such solution, it will contain 3.40 moles of the compound:

3.40 moles NaIO4 x 214 g/1 mol NaIO4 = 727.6 g

Besides, 1 L of solution equals 1090 g of the same solution, according to the density.

(1090 - 727.6) g = 362.4 g of H2O x 1 mol/18 g H2O = 20.13 moles

XNaIO4 = 3.40/(3.40 + 20.13) = 0.144

XH2O = 1 - 0.144 = 0.856

b. First, we calculate the moles of CoCl2 and NaOH reacting:

M = # moles/L solution ---> # moles = M * L solution

For CoCl2: 5.99x10-2 M * 0.317 L = 0.0189883 moles of CoCl2

For NaOH: 1.29x10-2 M * 0.864 L = 0.0111456 moles of NaOH

From the stoichiometry of the reaction, CoCl2/NaOH rate is 1/2, i.e. you need 2 moles of NaOH for every 1 mol of CoCl2. Therefore, NaOH is the reagent completely reacting, in the following manner:

0.0111456 moles of NaOH reacts with 0.0055728 moles of CoCl2 to produce 0.0111456 moles of Co(OH)2. Converting to mass:

0.0111456 moles of Co(OH)2 x 93 g/1 mol of Co(OH)2 = 1.0365 g of Co(OH)2 are produced.

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