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1

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
The number of moles of CuO, that will be utilized in Dumas method for estimation nitrogen in a sample of 57.5 g of N, N-dimethylaminopentane is _____________ $$\times$$ 10$$-$$2. (Nearest integer)

## Explanation

Moles of N in N, N - dimethylaminopentane = $$\left( {{{57.5} \over {115}}} \right)$$ = 0.5 mol

$$\Rightarrow {C_7}{H_{17}}N + {{45} \over 2}CuO \to 7C{O_2} + {{17} \over 2}{H_2}O + {1 \over 2}{N_2} + {{45} \over 2}Cu$$

$${{{n_{CuO}}\,reacted} \over {\left( {{{45} \over 2}} \right)}} = {{{n_{{C_7}{H_{17}}N}}\,reacted} \over 1}$$

$$\Rightarrow$$ nCuO reacted = $$\left( {{{45} \over 2}} \right) \times 0.5 = 11.25$$
2

### JEE Main 2021 (Online) 16th March Evening Shift

Numerical
In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is __________. (Round off to the Nearest Integer). [Given : Aqueous tension at 287 K = 14 mm of Hg]

## Explanation

Aqueous tension at 287 K = 14 mm of Hg.

Hence actual pressure = (758 – 14)
= 744 mm of Hg.

Moles of $${N_2} = {{\left( {758 - 14} \right)} \over {760}} \times {{30 \times {{10}^{ - 3}}} \over {0.0821 \times 287}}$$

$$= 1.246 \times {10^{ - 3}}$$ mol

mass of $${N_2} = 1.246 \times {10^{ - 3}} \times 28$$

mass % of N2 $$= {{mass\,of\,'N'} \over {total\,mass}} \times 100$$

$$= {{1.246 \times 28 \times {{10}^{ - 3}}} \over {0.184}} \times 100$$

$$= {{124.6 \times 28} \over {0.184}}\% = 18.96\%$$

$$\simeq 19\%$$
3

### JEE Main 2021 (Online) 25th February Morning Slot

Numerical
Using the provided information in the following paper chromatogram : Fig : Paper chromatography for compounds A and B.

the calculated Rf value of A ________$$\times$$ 10-1.

## Explanation

$${R_f} = {{Dis\tan ce\,travelled\,by\,compound} \over {Dis\tan ce\,travelled\,by\,solvent}}$$

On chromatogram distance travelled by compound is = 2 cm

Distance travelled by solvent = 5 cm

So, $${R_f} = {2 \over 5} =$$ 4 $$\times$$ 10$$-$$1 = 0.4
4

### JEE Main 2021 (Online) 24th February Evening Slot

Numerical
1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is __________ $$\times$$ 10$$-$$2.

## Explanation Given, weight = 18.6 g

Here, 1 mole of aniline gives 1 mole of acetanilide

$$\therefore$$ mole of aniline = mole of acetanilide

$$\Rightarrow$$ $${{1.86} \over {93}} = {{{W_{Acetanilide}}} \over {135}}$$

$${W_{Ace\tan ilide}} = {{1.86 \times 135} \over {93}}g = 2.70g$$

But efficiency of reaction is 90% only.

Hence, mass of acetanilide produced $$= 2.70 \times {{90} \over {100}}g = 2.43g = 243 \times {10^2}g$$

x = 243

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