18gram glucose dissolved in 500 gram water . Find the depression freezing point
Answers
Answered by
5
The freezing point to the solution is 272.8 K
Explanation:
To find out the freezing point, follow the procedure mentioned below
- Weight of water = m1 = 500 g
- Weight of glucose = m2 = 18 g
- Molar mass of water = M1 = 18 g/mole
- Molar mass of Glucose = M2 = 180 g/mole
- Freezing point depression constant for water = kf = 1.86 Kk g/ mole
Now, applying the formula, we get
ΔT = Kf * m2 * 1000/ M2 * m1
ΔT = 1.86 *18 *1000/ 180* 500
ΔT = 0.372
Then,
ΔT = Tf° - Tf
Tf = Tf° -ΔT
Tf = 273.15- 0.372 = 272.8 K
Answered by
5
Answer:
Explanation:
We know depression of freezing point is a colligative property.
∆T=Kf*m(where Kf is cryoscopic constant,m is morality)
Kf of water is 1.86°C/mm=No of moles/Mass of solvent in kg=(18/180)/(500/1000)=1/5
So, depression is 1.86*1/5=0.372°C
Similar questions
Math,
7 months ago
Computer Science,
7 months ago
English,
7 months ago
Social Sciences,
1 year ago
Geography,
1 year ago