CBSE 11TH CHEMISTRY - Online Test
Q1. To separate mixtures into their components, and also to test the purity of ompounds, the popularly used technique is :
Answer : Option A
Explaination / Solution:
Chromatography is used.
Q2.
A student forgot to add the reaction mixture to the round bottomed flask at 27 but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 . What fraction of air would have been expelled out?
Answer : Option A
Explaination / Solution:
n1/n2=T2/T1 =0.4 so fraction of air expelled out =1-0.6=0.4.
Q3. What will be the value of pH of
Answer : Option D
Explaination / Solution:
pH= -log(cα) And Ka=
Q4. The pH of ammonium acetate will be
Answer : Option A
Explaination / Solution:
is a salt of weak acid and weak base. pH= 7+ 0.5(pKa-pKb) pKa=-log(Ka) and pKb=-log(Kb) Here pKa=pKb so pH=7.
Q5.
The oxidation number of hydrogen in LiH, NaH and CaH7 is
Answer : Option C
Explaination / Solution:
This is according to the rules of assigning oxidation number i.e.
Metal hydrides, such as NaH, LiH, etc., in which the oxidation state for H is -1.
Q6. There are ___m in 2000 mm?
Answer : Option C
Explaination / Solution:
Since , 1 m
=1000 mm.
or, 1 mm
=
2000 mm
=[ ] m
= 2 m
Q7. Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of mono substituted alkyl halide?
Answer : Option D
Explaination / Solution:
(CH3)3CCH3 + HX(X=Cl,Br,I) ------------> (CH3)3CCH2X
Hence a monosubstituted derivative is formed.
Q8.
Enthalpy of combustion of carbon to is –393.5 kJ . Calculate the heat released upon formation of 35.2 g of from carbon and dioxygen gas.
Answer : Option A
Explaination / Solution:
when 1 mole of is produced energy released is –393.5 kJ Moles of given =35.2/44 =0.8 moles So energy released = 0.8 x393.5 KJ/mol = 315 KJ/mol
Q9. The elements charecterised by the filling of 4 f-orbitals, are:
Answer : Option A
Explaination / Solution:
The two rows of elements at the bottom of the Periodic Table, called the Lanthanoids, Ce(Z = 58) – Lu(Z = 71) and Actinoids, Th(Z = 90) – Lr (Z = 103) are characterised by the outer electronic configuration (n-2)f1-14 (n-1)d0-1ns2. The last electron added to each element is filled in f- orbital. These two series of elements are hence called the Inner-Transition Elements (f-Block Elements).
Q10. Around 1015 Hz corresponds to the region of the electromagnetic spectrum
Answer : Option C
Explaination / Solution:
Electromagnetic radiation in this range of wavelengths is called visible light or simply light. A typical human eye will respond to wavelengths from about 390 to 700 nm. In terms of frequency, this corresponds to a band in the vicinity of 430–770 THz.
