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Some Basic Concepts of Chemistry question

2021 · 31 Aug · Shift 1 · Q15
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Some Basic Concepts of Chemistry question

2021 · 31 Aug · Shift 1 · Q15

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H2C2O4.2H2OH_2C_2O_4.2H_2OH2​C2​O4​.2H2​O) in 250 mL of water in mol L −-− 1 is x ×\times× 10 −-− 2. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0]
Numerical answer
View written solutionFree

Correct answer: 20

  1. Write the formula and find molar mass

Given oxalic acid dihydrate: H2C2O4⋅2H2OH_2C_2O_4 \cdot 2H_2OH2​C2​O4​⋅2H2​O

Molar mass of H2C2O4H_2C_2O_4H2​C2​O4​: 2(1)+2(12)+4(16)=2+24+64=902(1) + 2(12) + 4(16) = 2 + 24 + 64 = 902(1)+2(12)+4(16)=2+24+64=90

Molar mass of 2H2O2H_2O2H2​O: 2(2(1)+16)=2(18)=362\big(2(1)+16\big)=2(18)=362(2(1)+16)=2(18)=36

So, total molar mass: 90+36=126 g mol−190+36=126\ \text{g mol}^{-1}90+36=126 g mol−1

  1. Calculate moles of solute

Mass given = 6.36.36.3 g

Moles=6.3126=0.05 mol\text{Moles} = \frac{6.3}{126} = 0.05\ \text{mol}Moles=1266.3​=0.05 mol

  1. Convert volume into litres

Volume of solution taken as 250250250 mL =0.250= 0.250=0.250 L

  1. Calculate molarity

M=molesvolume in L=0.050.25=0.20 mol L−1M = \frac{\text{moles}}{\text{volume in L}} = \frac{0.05}{0.25} = 0.20\ \text{mol L}^{-1}M=volume in Lmoles​=0.250.05​=0.20 mol L−1

  1. Match with given form

Given: Molarity=x×10−2\text{Molarity} = x \times 10^{-2}Molarity=x×10−2

Now, 0.20=20×10−20.20 = 20 \times 10^{-2}0.20=20×10−2

So, x=20x=20x=20

  1. Comparison with stored answer

Stored correct answer = 202020

Our derived answer matches it.

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