1

### JEE Main 2017 (Online) 8th April Morning Slot

Excess of NaOH (aq) was added to 100 mL of FeCl3 (aq) resulting into 2.14 g of Fe(OH)3 . The molarity of FeCl3 (aq) is :

(Given molar mass of Fe = 56 g mol−1 and molar mass of Cl = 35.5 g mol−1)
A
0.2 M
B
03 M
C
0.6 M
D
1.8 M

## Explanation

3 NaOH (aq.) + FeCl3(aq) $\to$ Fe(OH)3(s)+ 3 NaCl(aq).

Moles of Fe(OH)3 = ${{2.14} \over {107}}$ = 2 $\times$ 10$-$2

1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3

$\therefore\,\,\,$ 2 $\times$ 10$-$2 moles of Fe(OH)3 will obtain from

= 0.02 mole of FeCl3

Molarity of FeCl3 = ${{No.of\,moles} \over {Volume\,in\,L}}$ = ${{2 \times {{10}^{ - 2}}} \over {0.1}}$ = 0.2 M
2

### JEE Main 2017 (Online) 9th April Morning Slot

What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution ?
A
320
B
325
C
316
D
330
3

### JEE Main 2018 (Offline)

The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is
A
C2H4O3
B
C3H6O3
C
C2H4O
D
C3H4O2

## Explanation

$\therefore\,\,\,$ The ratio of no of atoms of C and H in one molecule of Cx Hy Oz = 1 : 2

$\therefore\,\,\,$ y = 2x

In one molecule of Cx Hy Oz compound contain z atoms of oxygen.

According to the question, no of atoms of oxygen required to burn CxHy completely should be twice of z atoms of oxygen.

CxHy + (x + ${y \over 4}$) O2 $\to$ HCO2 + ${y \over 2}$ H2O

No. of O2 molecules required = (x + ${y \over 4}$)

$\therefore\,\,\,$ No. of O atoms required = 2 (x + ${y \over 4}$)

According to question,

$2\left( {x + {y \over 4}} \right) = 2z$

$\Rightarrow \,\,\,2\left( {x + {{2x} \over 4}} \right) = 2z\,\,\,$ [ Putting y = 2x ]

$\Rightarrow \,\,\,\,{{3x} \over 2} = z$

$\therefore\,\,\,$ x : y : z

= x : 2x : ${{3x} \over 2}$

= 2 : 4 : 3

$\therefore\,\,\,$ Empirical formula = C2H4O3
4

### JEE Main 2018 (Online) 15th April Morning Slot

A sample of $NaCl{O_3}$ is converted by heat to $NaCl$ with a loss of $0.16$ $g$ of oxygen. The residue is dissolved in water and precipitated as $AgCl.$ The mass of $AgCl$ (in $g$) obtained will be : (Given : Molar mass of $AgCl=143.5$ $g$ $mo{l^{ - 1}}$)
A
$0.35$
B
$0.41$
C
$0.48$
D
$0.54$

## Explanation

NaClO3 $\buildrel \Delta \over \longrightarrow$ 2NaCl + 3O2

Here O2 produced = 0.16 g

$\therefore\,\,\,\,$ No of moles of O2 = ${{0.16} \over {32}}$ = 5 $\times$10$-$3

Let no of moles of NaCl

$\therefore\,\,\,\,$ ${{{}^nNaCl} \over 2}$ = ${{^n{O_2}} \over 3}$

$\Rightarrow$ $\,\,\,\,$ $^nNaCl$ = ${2 \over 3}$ $\times$ 5 $\times$ 10$-$3 = ${1 \over {300}}$

NaCl + Ag+ $\to$ AgCl + Na+

From here, you can see

no of moles in NaCl = no of moles in AgCl

$\therefore\,\,\,\,$ no of moles in AgCl = ${1 \over {300}}$

$\therefore\,\,\,\,$ Mass of AgCl = 143.5 $\times$ ${1 \over {300}}$ = 0.48 g