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For 109% labelled oleum if the number of moles of h2so4 and free so3 be x and y respectively,then approximate value of x y / x-y ?
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For 109% labelled oleum if the number of moles of h2so4 and free so3 b...
As we know that oleum actually a mixture of H2SO4 and  the free SO3
and in the 109% grams oleum there is actually 40 gram of SO3 and  also 60 gram of H2SO4 
So in this case, Number of moles of  H2SO4 will be "x'' and it will equal to Mass /molar mass (number of moles= Mass/ molar mass).
So the x will be = Mass/molar mass=  60 / 98 =   0.612 
Similarly number of moles of SO3 will be y and = mass/molar = 40/80 = 0.5=y
Now let's calculate the value of x+y = 0.612 + 0.5 = 1.112 Ans
Similarly x-y = 0.612 - 0.5= 0.112 Ans
Note ;molar mass of H2SO4 is 98 while for SO3 it is 80

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For 109% labelled oleum if the number of moles of h2so4 and free so3 b...
Oleum is mixture of H2SO4 and free SO3.

Labeling of oleum is based on the total amount of H2SO4 formed after reacting 100 g oleum with just enough H2O. The additional mass comes form amount of H2O required to convert all SO3 present in 100 g oleum into complete H2SO4.

Reaction involved is SO3 + H2O = H2SO4.

Here, 18 g H2O is required for 80 g SO3.

So, in 109 % oleum, 9 g H2O would be required for 40 g SO3.

It simply means that in 100 g oleum, 40 g (free) SO3 and 60 g H2SO4 are present.

So, in 100 g Oleum, number of moles of H2SO4 = (60/98) = 0.612 = X &

number of moles of SO3 = (40/80) = 0.5 = Y.

(X + Y) = 1.112

(X - Y) = 0.112

So, (X + Y)/ ( X-Y) = 1.112/ 0.112 = 9.92 = 10 (approx.)
Community Answer
For 109% labelled oleum if the number of moles of h2so4 and free so3 b...
Solution:

To find the approximate value of x y / x-y for 109% labelled oleum, we need to first understand the concept of oleum and its composition.

Oleum:
Oleum is a highly concentrated form of sulfuric acid (H2SO4). It is commonly known as fuming sulfuric acid or Nordhausen acid. Oleum is formed by dissolving sulfur trioxide (SO3) in sulfuric acid.

Composition of Oleum:
Oleum is represented by the formula H2SO4.xSO3, where x represents the amount of free SO3 present in the oleum. The value of x can vary depending on the concentration of the oleum.

Given Information:
In this case, we have 109% labelled oleum. This means that the oleum contains 109% of the required amount of SO3. Let's assume the number of moles of H2SO4 to be x and the number of moles of free SO3 to be y.

Calculating the Moles of H2SO4 and SO3:
Since oleum is a concentrated form of sulfuric acid, we can assume that the number of moles of H2SO4 is equal to the number of moles of SO3. Therefore, x = y.

Approximating x y / x-y:
We need to calculate the value of x y / x-y, which can be done as follows:

x y / x-y = (x * x) / (x - x)

As x and y are equal, the above expression simplifies to:

x y / x-y = (x * x) / 0

Since the denominator is zero, the value of x y / x-y is undefined.

Explanation:
The approximate value of x y / x-y for 109% labelled oleum is undefined because the number of moles of H2SO4 and free SO3 in oleum are assumed to be equal. If we assume any value for x and y, the denominator of the expression becomes zero, resulting in an undefined value.

Therefore, the approximate value of x y / x-y for 109% labelled oleum cannot be determined.

Conclusion:
The approximate value of x y / x-y for 109% labelled oleum is undefined. This is because the number of moles of H2SO4 and free SO3 in oleum are assumed to be equal, resulting in a zero denominator.
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