Ncert Solutions Chemistry Class 12th
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New answer posted
10 months agoContributor-Level 10
Given,
Volume of water, V = 450 mL = 0.45 L
Temperature, T= (37 + 273)K = 310 K
1.0 g of polymer of molar mass 185,000
Number of moles of polymer, n = 1 / 185,000 mol
We know that,
Osmotic pressure? = nRT/V
= 1 X 8.314 X 103 X 310 / 185000 X 0.45
= 30.98 Pa
= 31 Pa (approx)
New answer posted
10 months agoContributor-Level 10
Given
Mass of acetic acid, w1 = 75 g
Lowering of melting point? Tf = 1.5 K
Kf = 3.9 K kg/mol
Molar mass of ascorbic acid (C6H8O6), M2 6 * 12 + 8 * 1 + 6 * 16 = 176 g/mol
We know that,

= 5.08 g
Hence,
5.08 g of ascorbic acid is needed to be dissolved.
New answer posted
10 months agoContributor-Level 10
16.32

Alkyl groups are water hating groups that is they do not get assimilated in water easily whereas on the other hand functional groups like sulphonates, alcohol etc. are water liking groups that is these groups dissolve in water easily. Therefore hydrophobic part in the above detergents is the long chain alkyl group and the hydrophilic part is the alcohol group, sulphonates group etc.
New answer posted
10 months agoContributor-Level 10
Given,
Mass of water, wl = 500 g
Boiling point of water = 99.63°C (at 750 mm Hg).
Molal elevation constant, Kb = 0.52 K kg/mol
Molar mass of sucrose (C12H22O11), M2 (11 * 12 + 22 * 1 + 11 * 16) = 342 g/mol
Elevation of boiling point ΔTb = (100 + 273) - (99.63 + 273) = 0.37 K
We know that,
ΔTb = Kb X 1000 X W2 / M2 X W1
0.37 = 0.52 X 1000 X W2 / 340 X 500
w2 = 0.37 X 342 X 500 / 0.52 X 1000
w2 = 121.67 g
Hence,
121.67 g (approx) Sucrose is added to 500g of water so that it boils at 100°C.
New answer posted
10 months agoContributor-Level 10
Given,
Vapour pressure of water, PIo = 23.8 mm of Hg
Weight of water, w1 = 850 g
Weight of urea, w2 = 50 g
Molecular weight of water, M1 = 18 g/mol
Molecular weight of urea, M2 = 60 g/mol
n1 = w1/M1 = 850/18 = 47.22 mol
n2 = w2/M2 = 50/60 = 0.83 mol
We have to calculate vapour pressure of water in the solution p1
By using Raoult's therom,

PI = 23.4 mm of Hg Hence,
The vapour pressure of water in the solution is 23.4mm of Hg and its relative lowering is 0.0173.
New answer posted
10 months agoContributor-Level 10
Given, PAo = 450 mm Hg
PBo = 700 mm Hg
ptotal = 600 mm of Hg
By using Rault's law,
ptotal = PA + PB
ptotal = PAoxA + PBoxB
ptotal = PAoxA + PBo ( 1 - xA )
ptotal = (PAo- PBo)xA + PBo
600 = (450 - 700) xA + 700
-100 = -250 xA
xA = 0.4
∴ xB = 1 - xA
xB = 1 – 0.4
xB = 0.6
Now,
PA = PAoxA
PA = 450 * 0.4
PA = 180 mm of Hg and
PB = PBox
PB = 700 * 0.6
PB = 420 mm of Hg
Composition in vapour phase is calculated by
Mole fraction of liquid,
A =PA / PA + PB
= 180/180+420
= 0.30
Mole fraction of liquid,
B =PB / PA + PB
= 420 / 180+420
= 0.70
New answer posted
10 months agoContributor-Level 10
16.31
If calcium salts are present in water, then it is recommended to use synthetic detergent because soap will precipitate to give white scums and it will not clean the clothes.
On the other hand, if we use synthetic detergents, it will be dissolved in hard water also and provide effective cleaning of dirty clothes.
New answer posted
10 months agoContributor-Level 10
According to Henry's law, "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."
Stated as,
p = KHx
Where,
P = partial pressure of the solute above the solution
KH = Henry's constant
x = concentration of the solute in the solution
Given,
KH = 1.67x108 Pa
PCO2 2.5 atm = 2.5 * 1.01325 * 105Pa
According to Henry's law,
p = KHx
x = P/KH
x = 2.5 X 1.01325 X 105 / 1.67 X 108
x = 0.00152
In 500 ml of soda water there is 500 ml of water (neglecting soda)
Mole of water = 500/18
=
New answer posted
10 months agoContributor-Level 10
16.30 Soap molecules form micelles around an oil droplet [dirt] in such way that the hydrophobic parts of the stearate ions attach themselves to the oil droplet and the hydrophilic parts projects outside the oil droplet. Due to the polar nature of the hydrophilic parts, the stearate ions [along with the dirt] are pulled into the water, thereby removing the dirt from the cloth.

New answer posted
10 months agoContributor-Level 10
According to Henry's law,"At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."
Stated as,
p = KHx
Where, P = partial pressure of the solute above the solution
KH = Henry's constant
x = concentration of the solute in the solution
Given,
Solubility of H2S in water at STP is 0.195 m
We know,
At STP pressure p = 0.987 bar
0.195 mol of H2S is dissolved in 1000g of water
Moles of water = 1000/18
= 55.56 g/mol
∴ the mole fraction of H2S = Moles of H2S / Moles of H2S + Moles of water
&
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