1

### JEE Main 2018 (Online) 16th April Morning Slot

An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are :
(Atomic wt. of Cl = 35.5 u; Avogadro constant = 6.023 $\times$ 1023 mol-1)
A
6.023 $\times$ 1020
B
6.023 $\times$ 109
C
6.023 $\times$ 1021
D
6.023 $\times$ 1023

## Explanation

%    of   Cl  =  3.55

$\therefore\,\,\,\,$ In 100 g chlorohydrocarbon 3.55 gm Cl present.

In 1 gm chlorohydrocarbon Cl present

= ${{3.55} \over {100}}$

= 0.0355 gm

$\therefore\,\,\,\,$ No of Moles of Cl = ${{0.0355} \over {35.5}}$

= 0.001 mole

$\therefore\,\,\,\,$ no of Cl atoms = 0.001 $\times$ 6.023 $\times$ 103

= 6.023 $\times$ 1020
2

### JEE Main 2019 (Online) 9th January Morning Slot

Which amongst the following is the strongest acid ?
A
CHBr3
B
CHI3
C
CH(CN)3
D
CHCl3

## Explanation

Among those $-$CN has highest $-$M and $-$I effect. So after losing proton (H+) anion become more stable.

So,   CH(CN)3 has the highest tendency to donate proton that is why it is most acidic.
3

### JEE Main 2019 (Online) 9th January Morning Slot

A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in mol kg$-$1 is :
A
12
B
4
C
8
D
16

## Explanation

Molality of Na+ = ${{Moles\,\,of\,\,N{a^ + }} \over {mass\,\,of\,\,{H_2}O\,\,in\,\,kg}}$

Moles of Na+ = ${{92} \over {23}}$ = 4

Mass of H2O = 1 kg

$\therefore$   Molality = ${4 \over 1}$ = 4
4

### JEE Main 2019 (Online) 9th January Evening Slot

For the following reaction, in the mass of water produced from 445 g of C57H110O6 is :

2C57H110O6(s) + 163 O2(g) $\to$ 114 CO2(g) + 110 H2O(l)
A
490 g
B
445 g
C
495 g
D
890 g

## Explanation

moles of C57H110O6(s) = ${{445} \over {890}}$ = 0.5 moles

2C57H110O6(s) + 163 O2(g) $\to$ 114 CO2(g) + 110 H2O(l)

nH2O = ${{110} \over 4}$ = ${{55} \over 2}$

mH2O = ${{55} \over 2}$ $\times$ 18

= 495 gm