WBJEE 2015 Solved Chemistry Question Paper – Part 3

Find WBJEE 2015 Solved Chemistry Question Paper – Part 3 in this article. This paper consists of 5 questions (#51 to #55) from WBJEE 2015 Chemistry paper. Detailed solution of these questions has been provided so that students can match their solutions.

Importance of Previous Years’ Paper:

Previous years’ question papers help aspirants in understanding exam pattern, question format, important topics and assessing preparation. It has also been seen that sometimes questions are repeated in WBJEE Exam. So, this paper will certainly boost your confidence.

About WBJEE Exam

WBJEE is a common entrance examinations held at state level for admission to the Undergraduate Level Engineering and Medical courses in the State of West Bengal. The Chemistry section of WBJEE 2015 engineering entrance exam consists of 40 questions.

51. At a certain temperature, the value of the slope of the plot of osmotic pressure (∏) against concentration (C in mol L–1) of a certain polymer solution is 291R. The temperature at which osmotic pressure is measured is (R is gas constant)

(A) 271°C

(B) 18°C

(C) 564 K

(D) 18 K

Ans : (B)

Sol :

We have,

52. The rms velocity of CO gas molecules at 27°C is approximately 1000 m/s. For N2 molecules at 600 K the rms velocity is approximately

(A) 2000 m/s

(B) 1414 m/s

(C) 1000 m/s

(D) 1500 m/s

Ans : (B)

Sol:

The rms velocity of gas molecule is given as:

 53. A gas can be liquefied at temperature T and pressure P provided

(A) T = Tc and P < Pc

(B) T < Tc and P > Pc

(C) T > Tc and P > Pc

(D) T > Tc and P < Pc

Ans : (B)

Sol:

A gas can be liquified when T < TC and pressure P > PC

WBJEE 2016 Solved Physics and Chemistry Question Paper

54. The dispersed phase and dispersion medium of fog respectively are

(A) solid, liquid

(B) liquid, liquid

(C) liquid, gas

(D) gas, liquid

Ans : (C)

Sol:

The dispersed medium of fog is liquid and the dispersed phase is gas.

55. The decreasing order of basic character of K2O, BaO, CaO and MgO is

(A) K2O > BaO > CaO > MgO

(B) K2O > CaO > BaO > MgO

(C) MgO > BaO > CaO > K2O

(D) MgO > CaO > BaO > K2O

Ans : (A)

Sol:

The basicity of a compound depends on the electropositivity of metal.

So, the decreasing order of basic character is K2O > BaO > CaO > MgO.

Also Get:

WBJEE Sample Papers

WBJEE Previous Years' Question Papers

WBJEE Online Test

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