# Find molality of following two ions Mg2+ g/kg = 1.284 , mM =54.14HCO3- g/kg=0.126 , mM = 2.11 I need to find the molality of the ion. The density of seawater is 1.022 g/ml.

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### 1 Answer

Molality = moles of solute / kg of solvent

For Mg+2

If the concentration of Mg+2 = 54.14 mM,

There is 54.14 millimol Mg+2 per liter,

Which would be the same as 0.05414 mol/L.

Let's assume we have 1 L of seawater.

There would be 0.05414 mol of Mg+2 in the liter of sea water.

The mass of the liter of seawater would be...

From density...

1000 mL x 1.022 g/mL = 1022 g = 1.022 kg

Molality = 0.05414 mol / 1.022 kg

Molality = 0.0529 m or 5.29 x 10-2 m

For HCO3-

If the concentration of HCO3- = 2.11 mM,

There is 2.11 millimol HCO3- per liter,

Which would be the same as 0.00211 mol/L.

Let's assume we have 1 L of seawater.

There would be 0.00211 mol of HCO3- in the liter of sea water.

The mass of the sea water per liter would be...

From density...

1000 mL x 1.022 g/mL = 1022 g = 1.022 kg

Molality = 0.00211 mol / 1.022 kg

Molality = 0.00206 m or 2.06 x 10-3 m.