Advertisements
Advertisements
प्रश्न
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)
Advertisements
उत्तर
Kf for benzene = 5.12 K kg mol−1
WB = mass of solute = 25.6 g
WA = 1000 g
ΔTf = 5.12 K kg mol−1
Atomic mass of sulphur = 32 g mol−1
MB = `("K"_"f" xx "W"_"B" xx 1000)/(Δ"T"_"f" xx "W"_"A")`
= `(5.12 xx 25.6 xx 1000)/(0.512 xx 1000)`
= `131.072/0.152`
= 256
Now the molecular mass of
Sx = x × 32 = 256
x = `256/32` = 8
Therefore, the formula of sulphur = S8.
APPEARS IN
संबंधित प्रश्न
Write the formula to determine the molar mass of a solute using freezing point depresssion method.
1.0 x10-3Kg of urea when dissolved in 0.0985 Kg of a solvent, decreases freezing point of the solvent by 0.211 k. 1.6x10 Kg of another non-electrolyte solute when dissolved in 0.086 Kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)
Calculate the amount of CaCl2 (molar mass = 111 g mol−1) which must be added to 500 g of water to lower its freezing point by 2 K, assuming CaCl2 is completely dissociated. (Kf for water = 1.86 K kg mol−1)
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.
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.
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:
If molality of dilute solution is doubled, the value of molal depression constant (Kf) will be ______.
Latent heat of water (ice) is 1436.3 cal per mol. What will be molal freezing point depression constant of water (R = 2 cal/degree/mol)?
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
