1

### 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
2

### 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
3

### JEE Main 2019 (Online) 10th January Evening Slot

In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is :
A
10
B
2
C
1
D
5

## Explanation

$2\mathop M\limits^{ + 7} n{O_4} + 5{C_2}O_{_4}^{2 - } + 16{H^ + }\,\,\mathrel{\mathop{\kern0pt\longrightarrow} \limits_\,} \,2\mathop M\limits^{ + 2} {n^{2 + }}$

$+ 10C{O_2} + 8{H_2}O$

10 e$-$ trans for 10 molecules of CO2 so per molecule of CO2 transfer of e$-$ is '1'
4

### JEE Main 2019 (Online) 11th January Morning Slot

An organic compound is estimated through Duma's method and was found to evolve 6 moles of CO2. 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is :
A
C6H8N
B
C6H8N2
C
C12H8N
D
C12H8N2

## Explanation

Molar ratio of C : H : N : : 6 : 8 : 2 i.e., 3 : 4 : 1

Thus, the correct formula is C6H8N2. .