tiamariaazar
contestada

The following table lists molar concentrations of seven major ions in seawater. Using a density of 1.022 g/mL for seawater, convert the concentrations for the two ions in the question below into molality.

Ions g/kg mM
Na+ 10.781 480.57
K+ 0.399 10.46
Mg2+ 1.284 54.14
Ca2+ 0.4119 10.53
Cl- 19.353 559.40
SO42- 2.712 28.93
HCO3- 0.126 2.11
Total 35.067 n/a

Respuesta :

shinmin
Get the moles of the ions in the given above and the mass in kilograms of the pure seawater.

To get the mass, it is calculated by: total mass of all the present ions - mass of the solution

The mass of the solution:

1 L x 1.022 g /1 ml x 1 mL / 10^-3 L = 1022 g of solution

The total mass of the solute is given which is 35.067 g

The mass of pure solvent is: 1022 g - 35.067 g = 986.93 g = 0.98693 kg 

In this problem, the molality is given by: 10.53 mM = 0.01053 M = 0.01053 mol of

Ca^2+/L = 0.01053 mol / 0.98693 kg = 0.011 m is the answer.

Molarity of Ca²⁺ = 10.53 mM = 0.01053 M

Thus, moles of Ca²⁺ = Molarity × Volume

= 0.01053 M × 1 L

= 0.01053 moles

Density of solution = 1.022 g / ml

So, mass of solution = density × volume

= 1.022 g /ml × 1000 ml = 1022 g or 1.022 Kg

Mass of all ions in 1 Kg of seawater = 35.067 g

Mass of ions in 1.022 Kg = 1.022 Kg seawater × (35.067 g ions / 1 Kg seawater)

= 35.838 g

Therefore, mass of solvent, water = 1022 g - 35.838 g = 986.16 g or 0.98616 Kg

Molality of Ca²⁺ = moles / mass of solvent

= 0.01053 moles / 0.98616 Kg

= 0.0107 m

Similarly, molarity of HCO³⁻ = 2.11 mM or 0.00211 M

Consider a volume of 1 L (1000 ml) of solution

So, moles of HCO³⁻ = 0.00211 M × 1L = 0.00211 moles

So, mass of solvent, water = 1022 g - 35.838 g = 986.16 g or 0.98616 Kg

Molality of HCO³⁻ = moles of HCO³⁻ / mass of solvent

= 0.00211 moles / 0.98616 Kg

= 0.00214 m

Otras preguntas

Q&A Education