The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
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Answer:
315 g
Explanation:
Step 1: Write the thermochemical equation
2 H₂O(l) → 2 H₂(g) + O₂(g) ΔH = +572 kJ
Step 2: Calculate the molar of water decomposed by 5.00 × 10³ kJ of energy
According to the thermochemical equation, 572 kJ are required to decompose 2 moles of water.
5.00 × 10³ kJ × (2 mol/572 kJ) = 17.5 mol
Step 3: Calculate the mass corresponding to 17.5 moles of water
The molar mass of water is 18.02 g/mol.
17.5 mol × 18.02 g/mol = 315 g
3 Ca
2 P
8 O
There 13 mols of atoms.
