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JEE Main 202125 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Morning Shift
Subject
Chemistry
Chapter
Redox Reactions

JEE Main 2021 · 25 February · Morning Shift · Chemistry · Redox Reactions

0 . 4g mixture of NaOH, Na _ 2 CO _ 3 and some inert impurities was...

0.4g0 . 4g mixture of NaOH,Na2CO3NaOH, {Na}_{2}{CO}_{3} and some inert impurities was first titrated with N10HCl\dfrac{N}{10}HCl using phenolphthalein as an indicator, 17.5mL17 . 5mL of HClHCl was required at the end point. After this methyl orange was added and titrated. 1.5mL1 . 5mL of same HClHCl was required for the next end point. The weight percentage of Na2CO3{Na}_{2}{CO}_{3} in the mixture is (Rounded-off to the nearest integer)

Correct answer

0

Step-by-step solution

Upto first end point gm equi. of (NaOH+Na2CO3)=HCl\left(NaOH + {Na}_{2}{CO}_{3}\right) = HCl

x+y×1=110×17.5x + y \times 1 = \dfrac{1}{10} \times 17 . 5

x+y=1.75....(1)</p><p>Uptosecondendpoint</p><p>x + y = 1 . 75 . . . . \left(1\right)</p><p>Upto second end point</p><p>NaOH + {Na}_{2}{CO}_{3} ≡ HCl</p><p></p><p>x + y \times 2 = 110\dfrac{1}{10} \times 19</p><p></p><p>x + 2y = 1 . 9 . . . . \left(2\right)</p><p></p><p>y = 0 . 15</p><p></p><p>% {Na}_{2}{CO}_{3} = \dfrac{0 . 15 \times {10}^{- 3} \times 106}{0 . 4} \times 100</p><p></p><p>= 3 . 975 %</p><p></p><p>= 4 %</p><p>Henceansweris(4)</p><p>Hence answer is \left(4\right)

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