Freezing point depression depends of the number of particles of the solute in the solution.
1)Pure water have highest freezing point. All other solutions with given solutes will have lower temperatures.
2) The more particles of the solute in the solution the lower freezing point is going to be.
<span>b. 1.0 m NaCl ( dissociates and give 2 mol ions (1 mol Na⁺ and 1 mol Cl⁻))
c. 1.0 m K3PO4 (</span>dissociates and give 4 mol ions (3 mol K⁺ and 1 mol PO4³⁻)<span>
d. 1.0 m CaCl2 (</span>dissociates and give 3 mol ions (1 mol Ca²⁺ and 2 mol Cl⁻))<span>
e. 1.0 m glucose (c6h12o6) (glucose does not dissociate, and solution have
1 mole of particles of the solute(glucose))
The largest number of particles has </span>1.0 m K3PO4 solution, and it is has lowest freezing point . Answer is C.
C is correct. Have a good day!
If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
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What is base dissociation constant?
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The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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44g of CO2 can produce by the reaction of carbon with oxygen
