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Gaurav Seth 6 years, 8 months ago
Antacids are the substances that neutralise stomach acidity. Excessive secretion of acid for digestion in our stomach is fatal and can cause many diseases. So to neutralise it we use antacids.One well known antacid is Milk of Magnesia chemically known as magnesium Hydroxide. Second Example is Andrews Antacid. This is a type of Neutralisation Reaction.
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Gaurav Seth 6 years, 8 months ago
Phenol is more acidic than alcohols due to stabilisation of phenoxide ion through resonance. Presence of electron withdrawing group increases the acidity of phenol by , stabilising phenoxide ion while presence of electron releasing group decreases the acidity of phenol by destabilising phenoxide ion.
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Gaurav Seth 6 years, 8 months ago
Mole fraction of solution = solute + solvent = 1
Here, Mole fraction of water is 0.85 so mole fraction of sulphuric acid will be (1-0.85) = 0.15
Hence each mole of the solution, 0.15 mole of sulphuric acid is dissolved .
So mass of solvent (water) is ,
For 1 mole of water H2O = 18 gm
So, 0.85 mole of water = 18 x 0.85 = 15.3 gm =0.0153 Kg
Thus, Molality (m) = no. of moles of solute (Sulphuric acid)/ mass of solvent in Kg
Hence, Molality of the solution (m) = 0.15 / 0.0153 = 9.8 m (nearly equal to 10)
Posted by Divya Rawat 6 years, 8 months ago
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Gaurav Seth 6 years, 8 months ago
- In the presence of peroxides, the addition of HBr occurs by a free-radical mechanism and the orientation is anti-Markovnikov. This is true only for HBr.
- Free radical addition of HF and HI has never been observed, even in the presence of peroxides, and of HCl only rarely.
- In the rare cases where free-radical addition of HCl was noted, the orientation was still Markovnikov, because the more stable product was formed.
- Free radical addition of HF, HI and HCl is energetically unfavourable.
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Gaurav Seth 6 years, 8 months ago
1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water = 18 g mol - 1
∴ Number of moles present in 1000 g of water = 1000/18
= 55.56 mol
Therefore, mole fraction of the solute in the solution is
x2 = 1 / (1+55.56) = 0.0177.
It is given that,
Vapour pressure of water, p10 = 12.3 kPa
Applying the relation, (P10 - P1) / P10 = X2
⇒ (12.3 - p1) / 12.3 =0.0177
⇒ 12.3 - P1 = 0.2177
⇒ p1 = 12.0823
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.

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Appy Garnara 6 years, 8 months ago
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