Answer:
The freezing point of the solution is -1.4°C
Explanation:
Freezing point decreases by the addition of a solute to the original solvent, <em>freezing point depression formula is:</em>
ΔT = kf×m×i
<em>Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.</em>
Molality of the solution is:
-moles CaBr₂ (Molar mass:
189.9g ₓ (1mol / 199.89g) = 0.95 moles
Molality is:
0.95 moles CaBr₂ / 3.75kg water = <em>0.253m</em>
Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:
CaBr₂(s) → Ca²⁺ + 2Br⁻
3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.
Replacing:
ΔT = kf×m×i
ΔT = 1.86°C/m×0.253m×3
ΔT = 1.4°C
The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,
<h3>The freezing point of the solution is -1.4°C</h3>
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Answer:
I've already given an answer to your questions in your previous post. please check it
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<h2>Changes Occurs when a Barium Atom - Option 3 </h2>
When a barium atom loses two electrons it becomes a positive ion and its radius decreases. Barium (Ba) has atomic number 56 so it has 2 electrons in first shell of an atom to become stable according to duplet rule. Then other 52 electrons revolve in the shells according to octet rules.
Another 2 electrons are in the outermost shell. To become stable electrons lose to form barium ions (Ba+2). Hence, by losing 2 electrons the outermost shell will be diminished so its radius decreases and by losing electrons it becomes positive ions.
