the freezing pointof a 0.05 molal solution of a non electrolyte in water is
Asked by sarvaterahul
| 1st Jul, 2016,
08:25: AM
Expert Answer:
∆Tf = kf x m = 1.86 x 0.05 = 0.093°C
Freezing point = 0-0.093 = -0.093°C
∆Tf = kf x m = 1.86 x 0.05 = 0.093°C
Freezing point = 0-0.093 = -0.093°C
Answered by Vaibhav Chavan
| 1st Jul, 2016,
09:18: AM
Concept Videos
- Molal elevation constant for benzene is 2.5K/m. A solution of some organic substance in benzene boils at 0.126°C higher than benzene. What is the molality of solution?
- Boiling point of water at 750mm of hg is 99.63°C. how much sucrose is to be added to 500g of water such that it boils at 100°C.
- 2.00g of non-electolyte solute dissolved in 100g of benzene, lowered the freezing point of benzene by 0.30K. find the molar mass of solute.
- 3gm
- What is colligative property
- Explain B.pt.elevation and F.pt.depression?
- To 500 cm3of water ,3×10^-3 of acetic acid is added .if 23 percent is dissociated what will be the depression in freezing pt given k1 is 1.86K kg /mol
- two liquids a and b at 120C and 160C which of these liquids has at higher vapour pressure at 90C
- calculate freezing point of a solution when 54g of glucose is dissolved in 250g of water
- Calculate the amount of CaCl2, which must be added to 2 kg water so that the freezing point is depressed by 61 K (Kf of H2O =1.86 K kg mol^-1, Atomic mass of Ca = 40, Cl = 35.5)
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change