1 Stoichiometry
An aqueous solution containing 100grams of dissolved MgSO4 is fed to a crystallizer where 80% of the Dissolved salt crystallises out as MgSO4. 6H2O Crystals. How many grams of the hexahydrate salt Crystals are obtained from the crystalliser?(given Mg=24; S=32; O=16 and H=1).
1) 80. 2) 152. 3) 120. 4) 100
2 Basic Principles of Organic Chemistry
Identify the correct order of boiling points of the following compounds having nearly same molecular weight.
1) I > II > III > IV
2) IV > III > II > I
3) III > IV > II > I
4) IV > II > III > I
SOLUTION
Carboxylic acids have a higher boiling point than alcohol due to a more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding.
Alkane has the least boiling point due to the absence of any polarizing group.
So, the order is IV > III > II > I.
3 Haloalkanes and Haloarenes
The compound if contains an even number 'n' of chiral carbons, but the molecule can be divided into two equal and si,ilar halves , then how many optical active form it will have : 1) 2n 2) 2(n−1) 3) 2(n−1)/2 4) 2(n−1)−2(n−1)/2
4 Hydrocarbons
An alkyl bromide, RBr of molecular weight 151 is the exclusive product of bromination of which hydrocarbon?
1) Dodecane 2) 2,2 - dimethylpropane 3) 2,2 - dimethylhexane 4) 2,2,3 - trimethylheptane
Molecular weight of RBr=151
Molecular weight of Br=80
∴ Molecular weight of −R=151−80=71
If we consider 2,2−dimethylpropane the molecular formula is C5H12 .
After bromination, C5H11Br
∴ Molecular weight of C5H11=60+11=71
5 Hydrogen
In context with the industrial preparation of hydrogen from water gas (CO +H2), which of the following is the correct statement?
1) H2 is removed through occlusion with Pd 2) CO is oxidised to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in alkali 3) CO and H2 are fractionally separated using differences in their densities 4) CO is removed by absorption in aqueous Cu2C12 solution
The water gas is mixed with excess of steam and passed 45?C over a heated catalyst (Fe2O3+Cr2O3).CO is mostly oxidised to CO2 and more H2 is set free. CO2 is absorbed in alkali.
6 Surface Chemistry
Calculate the surface area in (sq m) of a catalyst that adsorbs 103cm3 of nitrogen reduced to STP per gram in order to form the monolayer. The effective area (in m2) occupied by N2? molecule on the surface is: ( surface area of N2?=1.62×10−15cm2)
1) 4115 m2 2) 2355 m2 3) 4355 m2 4) 435.5 m2
According to the given data,
Number of N2? molecules = Volume(ml) × NA / Volume at STP(in ml)
= (103×6.023×1023) / 22400 = 2.69×1022
Total area covered by N2? = 2.69×1022 × 1.62×10−15 = 435×105 cm2 = 4355 m2.
7 Alcohols Phenols and Ethers
Phenol is most easily soluble in 1. sodium carbonate solution 2. sodium hydroxide solution 3. dilute HCl 4. both sodium hydroxide and dilute HCl
8 Chemical Bonding and Molecular Structure
What type of hybridization take place in the N atom of NH3? 1) sp2 2) sp3 3) dsp2 4) sp
Hybridization Type sp3 Bond Angle 107o Geometry Pyramidal or Distorted Tetrahedral
9 Solid State
The edge length of the unit cell of NaCl crystal lattice is 552 pm. If the ionic radius of sodium ion is 95 pm., what is the ionic radius of chloride ion ?
1) 368 pm 2) 190 pm 3) 276 pm 4) 181 pm
10 Solid State
The number of atoms in 100 g of a fcc crystal with density of 10.0 g cm−3 and cell edge equal to 200 pm is equal to : 1) 5 × 1024 2) 5 × 1025 3) 6 × 1023 4) 2 × 1025
density $$ = \dfrac{zM}{N_a \times (a \times 10^{-10})^3}$$ $$10 = \dfrac{4 \times M}{6 \times 10^{23} \times (200\times 10^{-10})^3}$$ $$M= 12 \text{ g/mol}$$ Number of atoms is 100 g $$= \dfrac{100}{12} \times 6 \times 10^{23}= 5 \times 10^{24}$$
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