1

### 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. .
2

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

25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solutions ?
A
50 mL
B
12.5 mL
C
25 mL
D
75 mL

## Explanation

2HCl(aq) + Na2CO3(aq) $\to$ H2CO3 + NaCl

moles of HCl
2
=
moles of Na2CO3
1

$\Rightarrow$ ${{M \times {{25} \over {1000}}} \over 2} = {{0.1 \times {{30} \over {1000}}} \over 1}$

$\Rightarrow$ Molarity of HCl (M) = ${6 \over {25}}M$

HCl(aq) + NaOH(aq) $\to$ NaCl + H2O

moles of HCl
1
=
moles of NaOH
1

${{{6 \over {25}} \times {V \over {1000}}} \over 1} = {{0.2 \times {{30} \over {1000}}} \over 1}$

$\Rightarrow$ V = 25 ml
3

### JEE Main 2019 (Online) 12th January Morning Slot

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is -
A
20 g
B
4 g
C
80 g
D
10 g

## Explanation

Eq. of (COOH)2 = Eq. of NaOH

$\Rightarrow$ 50 $\times$ 0.5 $\times$ 2 = 25 $\times$ M $\times$ 1

$\Rightarrow$ M = 2 M

Now 1000 ml solution = 2 × 40 gram NaOH

$\therefore$ 50 ml solution = ${{2 \times 40 \times 50} \over {1000}}$ = 4 gram NaOH
4

### JEE Main 2019 (Online) 12th January Evening Slot

8 g of NaOH is dissolved in 18g of H2O. Mole fraction of NaOH in solution and molality (in mol kg–1) of the solution respectively are -
A
0.2, 11.11
B
0.167, 22.20
C
0.167, 11.11
D
0.2, 22.20

## Explanation

8 gm of NaOH   =   ${8 \over {40}}$   =   0.2 mol of NaOH

18 gm of H2O   =   ${18 \over {18}}$   =   1 mol of H2O

$\therefore$  Total mole   =   1 + 0.2   =   1.2 mol

$\therefore$  Mole fraction of NaOH   =   ${{0.2} \over {1.2}}$   =   0.167

We know,

Molality   =   ${{Moles\,\,of\,\,solute} \over {Weight\,of\,solvent}} \times$ 1000

=  ${{0.2} \over {18}} \times$ 1000

=  11.11