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Q.

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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a

-1.21

b

-1.51

c

1.51

d

1.21

answer is B.

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Detailed Solution

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 x+yx−y is approximately: