Test Preparation on IIT JEE Mains Mock Test-11

CHEMISTRY
An aqueous solution of glucose is 10% in strength. The volume in which 1 g mole of it is dissolved will be

118 lit
29 lit
30.9 lit
41.8 lit

10% solution means 10g in 100 mL;
1 mol of glucose = 180 g
10 g of glucose is present in 100 mL
180 g of glucose will be present in=100X180/10=1.8 Lit

Which of the following involve sp3d hybridisation?

1I3–
2IBr2–
3I5–
4ICI4–

Formation of I3- involves sp3d hybridization

Phospholipids are esters of glycerol with

1Three carboxylic acid residues
2Two carboxylic acid residues and one phosphate group
3One carboxylic acid residue and two phosphate groups
4Three phosphate groups

Strucure of a phospholipod is shown below

Which of the following molecules will not have a dipole moment?

1CH3Cl
2CH3OCH3
3CH2Cl2
4CCl4

The pair of radiation with lowest frequency, lowest wavelength respectively is represented by

1X–rays, microwaves
2X–rays, visible radiation
3-rays, microwaves
4microwaves, -rays

Microwave has lowest frerquency, γ-rays has
highest frequency (lowest wavelength).
Decreasing order of frequency: γ-rays > X-rays >
visible > microwave.

If the formula mass of Cr2O72– ion is M, then in the reaction
Cr2O72– + 14H⁺ → +6e– 2Cr3+ + 7H2O,
the equivalent mass of Cr₂O₇^2– is

1M
2M/2
3M/6
4M/14

In the given reaction oxidation state of Cr changes
from +6 → +3, i.e., 3 electrons per Cr atom. For
Cr2O7
2-, the oxidation state changes by 6, therefore
its equivalent wt = M/6

The formula of potassium dicyanobis(oxalato) nickelate (ll) is

1K4[Ni(CN)(Ox)2]
2K3[Ni2(CN)2(Ox)2]
3K4[Ni(CN)2(Ox)2]
4K2[Ni(CN)2(Ox)2]

The formula of potassium dicyanobis(oxalato)
nickelate (ll) isK4[Ni(CN)2(Ox)2]

The solubility product of BaCrO4 is 2.4 × 10–10 . The maximumconcentration of Ba(NO3)² possible without precipiitation in a 6 × 10–4 M K2CrO4 solution is

14 × 10–7 M
21.2 ×10–10 M
36 ×10–4 M
43 ×10–4 M

BaCrO4 ⇌ Ba2⁺+ CrO42-
Ksp = [Ba2+][CrO42-]
2.4 x 10-10= [Ba2+]x 6 x10-4 or
[Ba2+] = 0.4 x 10-6= 4 x 10-7

How many electrons are delivered at the cathode during electrolysis by a current of 1 amp.in 60 seconds?

16.0 × 1023
26.0 × 1020
33.74 × 1020
47.48 × 1021

Q = I x t = 1 x 60 =60 C
96500 C delivers 1 mol or 6.023 x1023 electrons at
cathode
60 C will deliver electrons = 6.023x10²³/96500 60=3.74 x 1020


Right Ans	Wrong Ans	Not Attempted

CHEMISTRYAn aqueous solution of glucose is 10% in strength. The volume in which 1 g mole of it is dissolved will be