Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12

Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) n-hexane and n-octane.
(ii) I₂ and CCl_4.(iii) NaCl O₄ and water.
(iv) methanol and acetone.
(v) acetonitrite (CH₃ CN) and acetone (C₃H₆O).

Answer:

(i) The intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B.
(ii) A-B interactions are weaker than those between A-A or B-B.
(iii) A-B interactions are more than those between A-A and B-B.
(iv) A-B interactions are less than A-A and B-B interactions.
(v) A-B interactions are more than A-A and B-B interactions.

208 Views

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molar solution of a non-volatile solute in it.

Answer:

1 molar solution means 1 number of moles of the solute per kilogram (kg) of the solvent. Mole fraction of solute.

= \frac{1}{1 + \frac{1000}{18}} = \frac{1}{1 + 55.55} = \frac{1}{56.55} \frac{P_{A}^{0} - P_{A}}{P_{A}^{0}} = x_{B}

\frac{12.3 - P_{A}}{12.3} = \frac{1}{56.55}

12.3 - P_{A} = \frac{12.3}{56.55} P_{A} = 12.3 - \frac{12.3}{56.55} = 12.3 - 0.2175 P_{A} = 12.0825 kpa = 12.08 kpa

216 Views

Amongst the following compunds identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform, (vi) pentanol.

Insoluble: toluene, chloroform.
Partially soluble: phenol, pentanol.
Highly soluble: formic acid, ethylene glycol.

166 Views

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain cyclohexane, KCl, CH₃OH, CH₃CN.

Answer:

n- octane is non-polar solvent therefore the solubility of a non-polar solute is more in n- octane solvent.

The order of increasing solubility is :

KCl<CH₃OH<CH₃CN< cyclohexane.

174 Views

Calculate the mass of a non-volatile solute (molecular mass 40 g mol^–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Answer:
molar mass of solute, M₂ =40g mol^-1mass of octane, w₁= 114g
molar mass of solute, M₁ = 114 g mol^-1

Let the vapour pressure of pure octane be

p_{A}^{0}

then, the vapour pressure of the octane after dissolving the non volatile solute is

\frac{80}{100} p_{1}^{0}

applying the relation,

\frac{P_{A}^{0} - P_{A}}{P_{A}^{0}} = x_{A}

\frac{100 - 80}{100} = \frac{\frac{ω_{B}}{M_{B}}}{\frac{ω_{A}}{M_{A}}}

\frac{20}{100} = \frac{ω_{B}}{40} \times \frac{114}{114}

ω_{B} = \frac{40 \times 20}{100} = 8 g

483 Views

Chapter Chosen

Book Chosen

Subject Chosen

Book Store

Previous Year Papers

Solutions

Calculate the mass of a non-volatile solute (molecular mass 40 g mol^–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Chapter Chosen

Book Chosen

Subject Chosen

Book Store

Previous Year Papers

Solutions

Calculate the mass of a non-volatile solute (molecular mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Calculate the mass of a non-volatile solute (molecular mass 40 g mol^–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.