Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or
)
Given : kf = 1.86 
m = 0.907
)
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86
* 0.907
)
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C
Answer:
Moles of NaCl formed is 6.0 moles
Explanation:
We are given the equation;
2 Na(s) + Cl₂(g) → 2 NaCl(s)
- Moles of Na is 6.0 moles
- Moles of Cl₂ is 4.0 moles
From the reaction;
2 moles of sodium reacts with 1 mole of chlorine gas to form 2 moles of NaCl
In this case;
6 moles of Na would require 3 moles of Cl₂, this means that chlorine gas is in excess.
Thus, the rate limiting reagent is sodium.
But, 2 moles of sodium reacts to form 2 moles of NaCl
Therefore;
Moles of NaCl = Moles of Na
= 6.0 moles
Thus, moles of sodium chloride produced is 6.0 moles
They depend on the electrons from an atom being distributed among them or shared, within their specialized bonding orbitals.
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Answer:
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