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1

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
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
2

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
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) 11th January Morning Slot

MCQ (Single Correct Answer)
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

JEE Main 2019 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)
English
Hindi
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

निम्न अभिक्रिया के लिए, $$445 \mathrm{g} \mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}$$ से उत्पादित जल का द्रव्यमान है:

$$\mathrm{2{C_{57}}{H_{110}}{O_6}(s) + 163{O_2}(g) \to 114C{O_2}(g) + 110{H_2}O(l)}$$

A
$$490 \mathrm{~g}$$
B
$$445 \mathrm{~g}$$
C
$$495 \mathrm{~g}$$
D
$$890 \mathrm{~g}$$

Questions Asked from Some Basic Concepts of Chemistry

On those following papers in MCQ (Single Correct Answer)
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