SolutionHard
Question
If 1 mole of a non-volatile and non-electrolyte solute in 1000 g of water depresses the freezing point by 1.86°C, then what will be the freezing point of a solution of 1 mole of the solute in 500 g of water?
Options
A.−0.93°C
B.−1.86°C
C.3.72°C
D.−3.72 °C
Solution
$\Delta T_{f} = K_{f}.m$
$1.86 = K_{f}.\frac{1}{1} $$$\Delta T_{f} = K_{f}.\frac{1}{0.5}$$
$\therefore\text{ Required }\Delta T_{f} = 3.72^{o}C \Rightarrow \text{ Freezing point = } - 3.72^{o}C$
Create a free account to view solution
View Solution FreeMore Solution Questions
A glucose solution is isopiestic with blood. What is the osmolarity of glucose solution?...The limiting value of van’t Hoff factor for Na2SO4 is...Aqueous solution of acedic acid contains :-...A liquid solvent is in equilibrium with its vapour. When a non-volatile solute is added to this liquid, the instant e ec...What will be the molecular weight of NaCl determined experimentally from elevation in the boiling point or depression in...