1

### JEE Advanced 2020 Paper 1 Offline

Numerical
5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette using phenolphthalein indicator. The volume of NaOH required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution?

Exp. No. Vol. of NaOH (mL)
1 12.5
2 10.5
3 9.0
4 9.0
5 9.0
Your Input ________

## Answer

Correct Answer is 0.11

## Explanation

Oxalic acid solution titrated with NaOH solution using phenolphthalein as an indicator.

$${H_2}{C_2}{O_4} + 2NaOH\buildrel {} \over \longrightarrow N{a_2}{C_2}{O_4} + 2{H_2}O$$

Equivalent of H2C2O4 reacted = Equivalent of NaOH reacted

$$= {{5 \times 2 \times 0.1} \over {1000}} = {{9 \times {M_{(NaOH)}} \times 1} \over {1000}}$$

$${M_{(NaOH)}} = {1 \over 9} = 0.11$$
2

### JEE Advanced 2020 Paper 2 Offline

Numerical
In the chemical reaction between stoichiometric quantities of KMnO4 and KI in weakly basic solution, what is the number of moles of I2 released for 4 moles of KMnO4 consumed?
Your Input ________

## Answer

Correct Answer is 6

## Explanation

In alkaline medium : Iodide is oxidised to iodate

$$2MnO_4^ - + {H_2}O + {I^ - }\buildrel {} \over \longrightarrow 2Mn{O_2} + 2O{H^ - } + IO_3^ -$$

But in weakly basic solution :

$$\mathop {KMn{O_4}}\limits^{ + 7} + \mathop {KI}\limits^{ - 1} \buildrel {} \over \longrightarrow \mathop {Mn{O_2}}\limits^{ + 4} + \mathop {{I_2}}\limits^0$$

Eq. of KMnO4 = Eq. of I2

$$4 \times 3 = n \times 2 \Rightarrow n = 6$$
3

### JEE Advanced 2019 Paper 2 Offline

Numerical
The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by conc. HNO3 to a compound with the highest oxidation state of sulphur is ..............

(Given data : Molar mass of water = 18 g mol$$-$$1)
Your Input ________

## Answer

Correct Answer is 288

## Explanation

When rhombic sulphur (S8) is oxidised by conc. HNO3 then H2SO4 is obtained and NO2 gas is released.

$${S_8} + 48HN{O_3}\buildrel {} \over \longrightarrow 8{H_2}S{O_4} + 48N{O_2} + 16{H_2}O$$

1 mole of rhombic sulphur produces = 16 moles of H2O

$$\therefore$$ Mass of water = 16 $$\times$$ 18 (molar mass of H2O) = 288 g
4

### JEE Advanced 2019 Paper 2 Offline

Numerical
The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm$$-$$3, then molarity of urea solution is ................

(Given data : Molar masses of urea and water are 60 g mol$$-$$1 and 18 g mol$$-$$1, respectively)
Your Input ________

## Answer

Correct Answer is 2.98

## Explanation

Molarity

$$(M) = {{Number\,of\,moles\,of\,solute \times 1000} \over {Volume\,of\,solution\,(in\,mL)}}$$

Also, volume = $${{Mass} \over {Density}}$$

Given, mole fraction of urea $$({\chi _{urea}})$$ = 0.05

Mass of water = 900 g

Density = 1.2 g/cm3

$${\chi _{urea}}$$ = $${{{n_{urea}}} \over {{n_{urea}} + 50}}$$

[$$\because$$ Moles of water = $${{900} \over {18}}$$ = 50]

$$0.05 =$$$${{{n_{urea}}} \over {{n_{urea}} + 50}}$$

$$\Rightarrow$$ $$19{n_{urea}} = 50$$

$${n_{urea}} = 2.6315$$ moles

$${w_{urea}} = {n_{urea}} \times {(M.wt)_{urea}}$$

$$= (2.6315 \times 60)g$$

$$V = {{2.6315 \times 60 + 900} \over {1.2}}$$

[$$\because$$ $$Density = {{Mass\,of\,solution} \over {Volume\,of\,solution}}$$]

= 881.57 mL

Now, molarity

= $$Number\,of\,moles\,of\,solute \times {{1000} \over {Volume\,of\,solution\,(mL)}}$$

$$= {{2.6315 \times 1000} \over {881.57}}$$ = 2.98 M

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