JEE Advanced Solved Chemistry Paper-2 Set X
Read this article to get free download of JEE Advanced 2017 Solved Physics Sample Paper-1 Set-X in pdf format.IIT JEE Examination is held to give admission to various prestigious engineering colleges such as IITs, NITs and various others prestigious engineering institutions. This examination is held at two levels: JEE Main and JEE Advanced.
IIT JEE Examination is held to give admission to various prestigious engineering colleges such as IITs, NITs and various others prestigious engineering institutions.
This examination is held at two levels: JEE Main and JEE Advanced
This JEE Advanced Solved Chemistry Paper-2 in this article consists of 20 questions divided into three sections. The questions in this paper have been asked from the complete syllabus. Each question of this paper is very important from examination point of view and has been developed very carefully by Subject Experts at Jagranjosh.
Few random questions from the sample paper are given below:
Q. Which of the following complex species is not expected to exhibit optical isomerism?
(a) [Co (en)3]3+
(b) Co (en)2Cl2]+
(c) [Co (NH3)3C13]
(d) [Co (en) (NH3 )C12 ]+
Correct Option: (b)
Optical isomerism is shown by those compounds which lacks elements of symmetry. Although [Co(NH3)3Cl3] shows facial as well as meridional isomerism. But both the isomeric forms contain plane of symmetry. So, [Co(NH3)3Cl3] does not show optical isomerism.
Paragraph based Questions
X and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure.
The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.
Q. The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is due to
(a) larger mean free path for X as a compared of that of Y
(b) larger mean free path for Y as compared to that of X
(c) increased collision frequency of Y with the inert gas as compared to that of X with the inert gas
(d) increased collision frequency of X with the inert gas as compared to that of Y with the inert gas
Ans. X is a lighter gas than Y, hence X has greater molecular speed. Due to greater molecular speed of X, it will have smaller mean free path and greater collision frequency with the inert gas molecules. As a result X will take more time to travel a given distance along a straight line. Hence X and Y will meet at a distance smaller than one calculated from Graham's law.
Hence, (d) is the correct choice.
Q. The value of d in cm (shown in the figure), as estimated from Graham's law, is
(a) 8 (b) 12 (c) 16 (d) 20
d = 48 - 2d
3d = 48
d = l6 cm
Hence, (c) is the correct choice