From other sources, the given mass of the solute that is being dissolved here is 7.15 g Na2CO3 - 10H2O. We use this amount to convert it to moles of Na2CO3 by converting it to moles using the molar mass then relating the ratio of the unhydrated salt with the number of water molecules. And by the dissociation of the unhydrated salt in the solution, we can calculate the moles of Na+ ions that are present in the solution.
Na2CO3 = 2Na+ + CO3^2-
7.15 g Na2CO3 - 10H2O (1 mol / 402.9319 g) (1 mol Na2CO3 / 1 mol Na2CO3 - 10H2O) ( 1 mol Na2CO3 / 1 mol Na2CO3-10H2O ) ( 2 mol Na+ / 1 mol Na2CO3) = 0.04 mol Na+ ions present
Answer:
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Some
of the solutions exhibit
colligative properties. These properties depend on the amount of solute
dissolved in a solvent. These properties include freezing point depression, boiling
point elevation, osmotic pressure and vapor pressure lowering. Calculations
are as follows:
<span>
ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>
