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
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)
Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol−1.
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:
The freezing point of equimolal aqueous solution will be highest for ____________.
Which has the highest freezing point?
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)?
Calculate freezing point depression expected for 0.0711 m aqueous solution of Na2SO4. If this solution actually freezes at – 0.320°C, what will be the value of van't Hoff factor? (kg for water = 108b°C mol–1)
1000 g of 1 m sucrose solution in water is cooled to −3.534°C. What weight of ice would be separated out at this temperature 1 is ______ gm. Kf(H2O) = 1.86 K mol−1 Kg)
40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is ______ K. (Nearest integer)
[Given : Kf = 1.86 K kg mol-1; Density of water = 1.00 g cm-3; Freezing point of water = 273.15 K]
Out of the following 1.0 M aqueous solution, which one will show the largest freezing point depression?
