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Some Basic Concepts of Chemistry question

2021 · 1 Sep · Shift 2 · Q19
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Some Basic Concepts of Chemistry question

2021 · 1 Sep · Shift 2 · Q19

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
If 80 g of copper sulphate CuSO4CuSO_4CuSO4​ . 5H2OH_2OH2​O is dissolved in deionised water to make 5L of solution. The concentration of the copper sulphate solution is x ×\times× 10 −-− 3 mol L −-− 1. The value of x is ‾\underline{\hspace{2cm}}​. [Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]
Numerical answer
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Correct answer: 64

  1. Find molar mass of CuSO4⋅5H2OCuSO_4\cdot 5H_2OCuSO4​⋅5H2​O
M(CuSO4⋅5H2O)=M(Cu)+M(S)+4M(O)+5(2M(H)+M(O))M(CuSO_4\cdot 5H_2O)=M(Cu)+M(S)+4M(O)+5\big(2M(H)+M(O)\big)M(CuSO4​⋅5H2​O)=M(Cu)+M(S)+4M(O)+5(2M(H)+M(O))

Substituting the given atomic masses:

=63.54+32+4(16)+5(2⋅1+16)=63.54+32+4(16)+5(2\cdot 1+16)=63.54+32+4(16)+5(2⋅1+16) =63.54+32+64+5(18)=63.54+32+64+5(18)=63.54+32+64+5(18) =63.54+32+64+90=249.54 g mol−1=63.54+32+64+90=249.54\ \text{g mol}^{-1}=63.54+32+64+90=249.54 g mol−1
  1. Calculate moles of CuSO4⋅5H2OCuSO_4\cdot 5H_2OCuSO4​⋅5H2​O

Given mass =80=80=80 g,

n=80249.54≈0.3206 moln=\frac{80}{249.54}\approx 0.3206\ \text{mol}n=249.5480​≈0.3206 mol
  1. Calculate molarity of the solution

Volume of solution =5=5=5 L,

C=nV=0.32065=0.06412 mol L−1C=\frac{n}{V}=\frac{0.3206}{5}=0.06412\ \text{mol L}^{-1}C=Vn​=50.3206​=0.06412 mol L−1
  1. Express in the form x×10−3 mol L−1x\times 10^{-3}\ \text{mol L}^{-1}x×10−3 mol L−1
0.06412=64.12×10−3 mol L−10.06412=64.12\times 10^{-3}\ \text{mol L}^{-1}0.06412=64.12×10−3 mol L−1

Hence,

x≈64x\approx 64x≈64

So, the required integer value is 64.

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