1
JEE Main 2018 (Offline)
+4
-1
For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?
A
[Co(H2O)3 Cl3].3H2O
B
[Co(H2O)6] Cl3
C
[Co(H2O)5 Cl] Cl2.H2O
D
[Co(H2O)4 Cl2] Cl.2H2O
2
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are Mx and My, respectively where Mx = $${3 \over 4}$$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :
A
$${4 \over 3}$$
B
$${3 \over 4}$$
C
$${1 \over 2}$$
D
$${1 \over 4}$$
3
JEE Main 2017 (Online) 9th April Morning Slot
+4
-1
A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol−1)
A
0.162
B
0.675
C
0.325
D
0.486
4
JEE Main 2017 (Online) 8th April Morning Slot
+4
-1
5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x

(Kf for water=1.86oC kg mol−1) is approximately :

(molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)
A
15 g
B
25 g
C
45 g
D
65 g
EXAM MAP
Medical
NEET