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:
-81.5 degrees C or 191.5 K
Explanation:
We want to use Charles' gas law: V/T = V/T
Our initial volume is 3.20 L, and our initial temperature is 125 degrees C, or 125 + 273 = 398 degrees Kelvin.
Our new Volume is 1.54 L, but we don't know what the temperature is. So, we use the equation:
3.20 L / 398 K = 1.54 L / T ⇒ Solving for T, we get: T = 191.5 K
If we want this in degrees Celsius, we subtract 273: 191.5 - 273 = -81.5 degrees C
A scale and a ruler. The scale to measure the mass, and a ruler to measure the volume.
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5