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Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Answer:
Vapour pressure of water pA0 = 17.535mm of Hg
Mass of glucose W2 = 25 g
mass of water w1 = 450g
we know that 
mass of glucose = 180gmol-1
molar mass of glucose =25/180g mol-1
mass of water = 18 g
molar mass of water M2 = 450/18 g mol-1
apply equation we get,

pA0-pApA0=xA


 17.535-pA17.535=25/180450/18


or      17.535-pA17.535 = 25180×18450=1180


or  180(17.535-pA) = 17.535

or  3156.30-180pA = 17.535

or                   3138.765 = 180pA

or                      pA = 3138.765180 = 17.44 mm Hg.
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100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B and found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Solution
Given that
Mass of liquid A , WA = 100g
Molar mass, MA = 140 g mol – 1
Mass of liquid B, WB = 1000 g
Molar mass, MB = 180 g mol – 1
Use the formula 
 Number of moles  = massmolar mass
 
Number of moles of liquid A, MA = 100/140 = 0.714 mol
Number of moles of liquid B, MB =1000/ 180 = 5.556 mol
Use formula 
 Mole fraction of A = number of molestotal number of moles
 
Molar fraction of A,XA = 0.714 /(0.714 + 5.556) = 0.114
Similarly
Molar fraction of B, XB   = 1-  XA
= 1 − 0.114 = 0.886
Vapour pressure of pure liquid B, PoB = 500 torr
Use formula of Henry`s law
PB  = Po× XB
Plug the values we get
PB= 500 × 0.886 = 443 torr
Given that total vapour pressure of the solution, ptotal = 475 torr
Use the formula
Ptotal   = pA + pB
pA      = ptotal − pB
Plug the values we get
PA      = 475 − 443
PA      = 32 torr
Use formula of Henry`s law again we get
PA  = Po× XA
Plug the values we get
32 = PoA × 0.114
PoA = 32/0.114  = 280.7 torr
So that the vapour pressure of pure liquid A = 280.7 torr.

 
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Henry's law constant for the molality of methane in benzene at 298 K is 4.27 x 105mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg. 

Answer:

p = 760mmHg
KH = 4.27 x 105

According to Henry's law
v = KHx
kH = 4.27 x 105 mm Hg,
p = 760 mm
                      760 = 4.27×105×x

or        x=7604.27×105 = 178×10-5.
315 Views

Nalorphene (C19 H21 NO3) similar to morphine, is used to combat withdrawl symptoms in narcotic users. Dose of nalophene generally given is 1.5 mg. Calculate the mass of 1.5 x10–3 m  aqueous solution required for the above dose.

Molar Mass of C19 H12 NO3
= 19 x 12 + 21 x 1 + 14 + 48
= 228 + 21 + 14 + 48
= 311 g mol–1
Molality (m) = Mass of solute/molarmassMass of solvant in kg1.5 = Mass of solute/31111or  1.5 = Mass of solute311or Mass of solute = 1.5×10-3 × 311 = 0.4665g

  Mass of total solution = 1000 g + 0.467 g  = 1000.467 g
To convert 1.5 g of nalorphene in to g divide by 1000 we get
1.5 mg / 1000  = 0.0015 g
Thus, 0.467 g mass of nalorphene contain

by      = 1000.467 g solution

 for 0.0015 g mass of nalorphene contain

by       = 1000.467 × 0.0015/0.467
 
= 3.21 g.

 

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Calculate the depression in the freezing of water when 10 g of CH3CH2CHCICOOH is added to 250 g of water. Ka = 1.4 x 10–3, K= 1.86 K kg moL–1


Answer:

We have given 
w2 = 10g
w1 = 250
Kf = 1.86k kg mol-1
M2 = 122.5
Mol. mass of CH3CH2CHCICOOH
     
=12+3+12+2+12+1×35.5+12+16+6+1= 122.5 g mol-1Tf = Kf×ω2×1000M2×ω1        = 1.86×10×1000122.5×250=  0.61°

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