SolutionHard
Question
x mole of KCI and y mole of BaCl2 are both dissolved in 1 kg of water. Given that x + y = 0.1 and Kf for water is 1.85 K/molal, what is the observed range of ᐃTf, if the ratio of x to y is varied ?
Options
A.0.37o to 0.505o
B.0.185o to 0.93o
C.0.56o to 0.93o
D.0.37o to 0.93o
Solution
x + y = 0.1
x / y = 0.37
ᐃTf =
× 1000 × 1.85
ᐃTf =
× 1000 × 1.85
ᐃTf = (0.2 + 0.0729927) × 1.85 = 0.505
x / y = 0.37
ᐃTf =
ᐃTf =
ᐃTf = (0.2 + 0.0729927) × 1.85 = 0.505
Create a free account to view solution
View Solution FreeMore Solution Questions
Depression of freezing point of 0.01 molal aq. CH3COOH solution is 0.02046o. 1 molal urea solution freezes at - 1.86°C. ...A liquid solvent is in equilibrium with its vapour. When a non-volatile solute is added to this liquid, the instant e ec...Two aqueous solutions S2 and S2 are separated by a semipermeable membrane. S1 has lower vapour pressure than S2. Which o...' W ' g of a non-volatile electrolyte solid solute of molar mass ' M ' $g{mol}^{- 1}$ when dissolved in 100 mL water, de...The boiling point of pure benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, th...