Advertisements
Advertisements
प्रश्न
Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol−1. Calculate the atomic masses of A and B.
उत्तर
We know that,
M2 = `(1000xx"w"_2xx"K"_"f")/(triangle"T"_"f"xx"w"_1)`
Then `"M"_("AB"_2) = (1000xx1xx5.1)/(2.3 xx 20)`
= 110.87 g mol−1
`"M"_("AB"_4) = (1000xx 1xx5.1)/(1.3xx20)`
= 196.15 g mol−1
Now, the molar masses of AB2 and AB4 are 110.87 g mol−1 and 196.15 g mol−1, respectively.
Let the atomic masses of A and B be x and y, respectively.
Now, we can write:
x + 2y = 110.87 ....(i)
x + 4y = 196.15 ....(ii)
Subtracting equation (i) from (ii), we have
2y = 85.28
⇒ y = 42.64
Putting the value of ‘y’ in equation (i), we have
x + 2 × 42.64 = 110.87
⇒ x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u, respectively.
APPEARS IN
संबंधित प्रश्न
Define Cryoscopic constant.
Which of the following solutions shows maximum depression in freezing point?
(A) 0.5 M Li2SQ4
(B) 1 M NaCl
(C) 0.5 M A12(SO4)3
(D) 0.5 MBaC12
Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water.
(Kf for water = 1.86 K kg mol–1)
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10−3, Kf = 1.86 K kg mol−1.
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water. (Kf of water = 1.86 K kg mol–1)
Give reasons for the following:
Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
A 4% solution(w/w) of sucrose (M = 342 g mol−1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol−1) in water.
(Given: Freezing point of pure water = 273.15 K)
Cryoscopic constant of a liquid is ____________.
Pure benzene freezes at 5.45°C. A 0.374 m solution of tetrachloroethane in benzene freezes at 3.55°C. The Kf for benzene is:
A solution containing 1.8 g of a compound (empirical formula CH2O) in 40 g of water is observed to freeze at –0.465° C. The molecular formula of the compound is (Kf of water = 1.86 kg K mol–1):
Which has the highest freezing point?
Of the following four aqueous solutions, total number of those solutions whose freezing points is lower than that of 0.10 M C2H5OH is ______. (Integer answer)
- 0.10 M Ba3 (PO4)2
- 0.10 M Na2 SO4
- 0.10 M KCl
- 0.10 M Li3 PO4
When 25.6 g of sulphur was dissolved in 1000 g of benzene, the freezing point lowered by 0.512 K. Calculate the formula of sulphur (Sr).
(Kf for benzene = 5.12 K kg mol−1, Atomic mass of sulphur = 32 g mol−1)
Out of the following 1.0 M aqueous solution, which one will show the largest freezing point depression?
Ibrahim collected 10 mL each of fresh water and ocean water. He observed that one sample labeled “P” froze at 0° C while the other “Q” at -1.3° C. Ibrahim forgot which of the two, “P” or “Q” was ocean water. Help him identify which container contains ocean water, giving rationalization for your answer.
The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid, and trifluoroacetic acid increases in the order given above. Explain briefly.
The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
