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Redox Reactions question

2020 · 4 Sep · Shift 2 · Q8
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Redox Reactions question

2020 · 4 Sep · Shift 2 · Q8

JEE MainChemistryRedox ReactionsNumerical+4 / −1
A 100 mL solution was made by adding 1.43 g of Na2CO3Na_2CO_3Na2​CO3​.xH2OxH_2OxH2​O. The normality of the solution is 0.1 N. The value of x is ‾\underline{\hspace{2cm}}​. (The atomic mass of NaNaNa is 23 g/mol)
Numerical answer
View written solutionFree

Correct answer: 10

  1. Find the equivalent weight of Na2CO3⋅xH2ONa_2CO_3 \cdot xH_2ONa2​CO3​⋅xH2​O

For acid-base reactions, Na2CO3Na_2CO_3Na2​CO3​ has valency factor n=2n = 2n=2 because it can neutralize 2H+2H^+2H+.

So, equivalent weight of hydrated sodium carbonate is

molar mass2=(106+18x)2\frac{\text{molar mass}}{2} = \frac{(106 + 18x)}{2}2molar mass​=2(106+18x)​

Here,

M(Na2CO3)=2(23)+12+3(16)=46+12+48=106M(Na_2CO_3) = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106M(Na2​CO3​)=2(23)+12+3(16)=46+12+48=106
  1. Use normality relation

Given:

  • Mass of solute =1.43 g= 1.43\,g=1.43g
  • Volume of solution =100 mL=0.1 L= 100\,mL = 0.1\,L=100mL=0.1L
  • Normality =0.1 N= 0.1\,N=0.1N

Number of equivalents in solution:

equivalents=N×V=0.1×0.1=0.01\text{equivalents} = N \times V = 0.1 \times 0.1 = 0.01equivalents=N×V=0.1×0.1=0.01

Hence,

equivalent weight=massequivalents=1.430.01=143\text{equivalent weight} = \frac{\text{mass}}{\text{equivalents}} = \frac{1.43}{0.01} = 143equivalent weight=equivalentsmass​=0.011.43​=143
  1. Equate the two expressions for equivalent weight
106+18x2=143\frac{106 + 18x}{2} = 1432106+18x​=143 106+18x=286106 + 18x = 286106+18x=286 18x=18018x = 18018x=180 x=10x = 10x=10
  1. Final answer
10\boxed{10}10​

The derived answer matches the stored correct answer.

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