SolutionHard

Question

The mass of KCl required to depress the freezing point of 500 g water by 2 K is (K_f = 1.86, K = 39)

Options

A.10.01 g

B.40.05 g

C.7.45 g

D.20.03 g

Solution

$\Delta T_{f} = K_{f}.m \Rightarrow 2 = 1.86 \times \frac{(w/74.5) \times 2}{500/1000} \Rightarrow w = 20.03\text{ gm}$

Create a free account to view solution

View Solution Free

Topic: Solution·Practice all Solution questions

More Solution Questions

When an ideal binary solution is in equilibrium with its vapour, molar ratio of the two components in the solution and i...Benzene and toluene form nearly ideal solutions. At 20oC, the vapour pressure of benzene is 75 torr and that of toluene ...The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile non-electrolyte solid weighing...For a dilute solution having molality m of a given solute in a solvent of mol.wt. M, b.pt. Tb and heat of vaporisation p...Phenol dimerizes in benzene. If the observed molecular mass of phenol in solution is 120, then its degree of dimerizatio...