For 118% labelled oleum if the no. of moles of H2SO4 and free SO3 be x and y respectively, the values of x+yx−y is approximately:

118% Oleum18 g water - 1 mol water1 mol SO3 = 80 g SO3∴ y=1∴ nH2SO4 in oleum (x)=20/98∴ x+yx−y=1+20982098−1=−1.51

For 118% labelled oleum if the no. of moles of H2SO4 and free SO3 be x and y respectively, the values of x+yx−y is approximately:

118% Oleum18 g water - 1 mol water1 mol SO3 = 80 g SO3∴ y=1∴ nH2SO4 in oleum (x)=20/98∴ x+yx−y=1+20982098−1=−1.51

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For 118% labelled oleum if the no. of moles of H₂SO₄ and free SO₃ be x and y respectively, the values of $(\frac{x + y}{x - y})$ is approximately:

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1.21

1.51

-1.51

-1.21

answer is B.

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Detailed Solution

118% Oleum
18 g water - 1 mol water

1 mol SO₃ = 80 g SO₃

$\begin{array}{ll} ∴ y = 1 \\ ∴ n_{H_{2} {SO}_{4}} {in oleum}_{(x)} = 20 / 98 \\ ∴ \frac{x + y}{x - y} = \frac{1 + \frac{20}{98}}{\frac{20}{98} - 1} = - 1.51 \end{array}$

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For 118% labelled oleum if the no. of moles of H2SO4 and free SO3 be x and y respectively, the values of x+yx−y is approximately:

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For 118% labelled oleum if the no. of moles of H₂SO₄ and free SO₃ be x and y respectively, the values of $(\frac{x + y}{x - y})$ is approximately: