1
IIT-JEE 2008 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-0

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.

A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.

Given:

Freezing point depression constant of water $$\left( {K_f^{water}} \right) = 1.86$$ K kg mol$$^{-1}$$

Freezing point depression constant of ethanol $$\left( {K_f^{ethanol}} \right) = 2.0$$ K kg mol$$^{-1}$$

Boiling point elevation constant of water $$\left( {K_b^{water}} \right) = 0.52$$ K kg mol$$^{-1}$$

Boiling point elevation constant of ethanol $$\left( {K_b^{ethanol}} \right) = 1.2$$ K kg mol$$^{-1}$$

Standard freezing point of water = 273 K

Standard freezing point of ethanol = 155.7 K

Standard boiling point of water = 373 K

Standard boiling point of ethanol = 351.5 K

Vapour pressure of pure water = 32.8 mm Hg

Vapour pressure of pure ethanol = 40 mm Hg

Molecular weight of water = 18 g mol$$^{-1}$$

Molecular weight of ethanol = 46 g mol$$^{-1}$$

In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.

The vapour pressure of the solution M is :

A
39.3 mm Hg
B
36.0 mm Hg
C
29.5 mm Hg
D
28.8 mm Hg
2
IIT-JEE 2008 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-0

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.

A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.

Given:

Freezing point depression constant of water $$\left( {K_f^{water}} \right) = 1.86$$ K kg mol$$^{-1}$$

Freezing point depression constant of ethanol $$\left( {K_f^{ethanol}} \right) = 2.0$$ K kg mol$$^{-1}$$

Boiling point elevation constant of water $$\left( {K_b^{water}} \right) = 0.52$$ K kg mol$$^{-1}$$

Boiling point elevation constant of ethanol $$\left( {K_b^{ethanol}} \right) = 1.2$$ K kg mol$$^{-1}$$

Standard freezing point of water = 273 K

Standard freezing point of ethanol = 155.7 K

Standard boiling point of water = 373 K

Standard boiling point of ethanol = 351.5 K

Vapour pressure of pure water = 32.8 mm Hg

Vapour pressure of pure water = 40 mm Hg

Molecular weight of water = 18 g mol$$^{-1}$$

Molecular weight of ethanol = 46 g mol$$^{-1}$$

In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.

Water is added to the solution M such that the fraction of water in the solution becomes 0.9 mole. The boiling point of this solution is:

A
380.4 K
B
376.2 K
C
375.5 K
D
354.7 K
3
IIT-JEE 2008 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1
The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors $$\overrightarrow a \,,\,\overrightarrow b ,\overrightarrow c $$ such that $$\widehat a\,.\,\widehat b = \widehat b\,.\,\widehat c = \widehat c\,.\,\widehat a = {1 \over 2}.$$ Then, the volume of the parallelopiped is :
A
$${1 \over {\sqrt 2 }}$$
B
$${1 \over {2\sqrt 2 }}$$
C
$${{\sqrt 3 } \over 2}$$
D
$${1 \over {\sqrt 3 }}$$
4
IIT-JEE 2008 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1

Let A, B, C be three sets of complex numbers as defined below :

$$A = \left\{ {z:\,{\mathop{\rm Im}\nolimits} \,\,z\,\, \ge \,1} \right\}$$

$$B = \left\{ {z:\,\,\left| {z - 2 - i} \right| = 3} \right\}$$

$$C = \left\{ {z:\,{\mathop{\rm Re}\nolimits} (1 - i)z) = \sqrt 2 \,} \right\}$$

The number of elements in the set $$A \cap B \cap C$$ is

A
0
B
1
C
2
D
$$\infty $$
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