0.1 M KI and 0.2 M `AgNO_(3)` are mixed in 3 : 1 volume ratio. The depression of freezing point of the resulting solution will be [`K_(b)(H_(2)O) = 1.86 K kg "mol"^(-1)`]:
A
3.72 K
B
1.86 K
C
0.93 K
D
0.279 K
Text Solution
Verified by Experts
The correct Answer is:
d
Topper's Solved these Questions
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 2|1 Videos
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
0.1M KI and 0.2 M AgNO_(3) are mixed in 3:1 volume ratio . The deperession in freezing point of the resulting solution will be 0.1 x (Assume Kf of H_(2)O=2Kg//"mol" ).
Equal volumes of 1.0M AgNO_(3) and 1.0M KCl are mixed . The depression of freezing point of the resulting solution will be (K_(f)(H_(2)O)=1.86 K kg "mol"^(-1), 1M=1m)
Mole fraction of solute in some solution is 1//n . If 50% of solute molecules dissociate into two parts and remaining 50% get dimerised , new mole fraction of solvent becomes 4//5 . Find value of n.29. 0.1 M KI and 0.2 M are mixed in 3:1 volume ratio . The depression in freezing point of the resulting solution will be 0.1 x (Assume K_(f) of and molality =molarity). Then x=_________
100 ml each of 1M AgNO_3 and 1M NaCl are mixed. The nitrate ion concentration in the resulting solution is
1 kg of 2m urea solution is mixed with 2 kg of 4m urea solution. The molality of the resulting solution is
A solution of sucrose (molar mass 342 "gmol"^(-1) ) has been produced by dissolving 68.5g sucrose in 1000g water. The freezing point of the solution obtained will be ( K_f for H_2O = 1.86 kg "mol"^(-1) )
1.0 molal aqueous solution of an electrolyte X_(3)Y_(2) is 25% ionized. The Boiling point of the solution is (K_(b) of H_(2)O=0.52K kg mol^(-1))
12.25 g of CH_(3)CH_(2)CHCICOOH is added to 250 g of water to make a solution . If the dissociation constant of above acid is 1.44xx10^(-3) the depression in freezing point of water in ""^(@)C "is" (K_(f) for water is 1.86 K kg "mol"^(-1))
NARENDRA AWASTHI-DILUTE SOLUTION-Level 3 - Match The Column
