1
MCQ (Single Correct Answer)

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
2
MCQ (Single Correct Answer)

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'
3
MCQ (Single Correct Answer)

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. .
4
MCQ (Single Correct Answer)

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

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