Fill in the Blanks
1. Given that ΔT_{f} is the depression in freezing point of the solvent in a solution of a nonvolatile solute of molality, m, the quantity is equal to ........... . (1994  1 Mark)
Ans: K_{f}
Solutions : K_{f}; Depression in freezing point, ΔT_{f} = K_{f} . m, where K_{f} is the molar depression constant or cryoscopic constant and m is the molality of the solution given by moles of solute per 1000 g of the solvent.
Integer Value Correct Type
1. 29.2% (w/w) HCl stock solution has a density of 1.25 g mL^{–1}. The molecular weight of HCl is 36.5 g mol^{–1}. The volume (mL) of stock solution required to prepare a 200 mL
solution of 0.4 M HCl is : (2012)
Ans : 8
Solution: Molarity of stock solution of HCl
Let the volume of stock solution required = V mL
Thus
2. MX_{2} dissociates into M^{2+} and X^{–} ions in an aqueous solution, with a degree of dissociation (a) of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is (JEE Adv. 2014)
Ans : 2
Solutions:
3. A compound H_{2}X with molar weight of 80g is dissolved in a solvent having density of 0.4 g ml^{–1}. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is (JEE Adv. 2014)
Ans : 8
Solutions :
Calculation of wt. of solvent
1 mL of solvent = 0.4 g
1000 mL of solvent = 400 g
Calculation of wt. of solute
1000 mL of solution contain = 3.2 × 80 g solute = 256 g
4. If the freezing point of a 0.01 molal aqueous solution of a cobalt(III) chlorideammonia complex (which behaves as a strong electrolyte) is –0.0558°C, the number of chloride(s) in the coordination sphere of the complex is [K_{f} of water = 1.86 K kg mol^{–1}] (JEE Adv. 2015)
Ans : 1
Solution :
Given ΔT_{f} = 0.0558°C
as we know, ΔT_{f} = i × K_{f} × m
⇒ 0.0558 = i × 1.86 × 0.01
i = 3
Therefore the complex will be [Co(NH_{3})_{5}Cl]Cl_{2}
Hence number of chloride in coordination sphere is 1.
5. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm^{–3} . The ratio of the molecular weights of the solute and solvent, , is
Ans : 9
solution : 1 mole solution has 0.1 mole solute and 0.9 mole solvent.
Let M_{1} = Molar mass solute
M_{2} = Molar mass solvent
Molality, m = .....(1)
Molarity, M = .......(2)
∵ m = M
142 docs66 tests

142 docs66 tests


Explore Courses for JEE exam
