Mole ConceptHard

Question

In 200 g of a sample of oleum labelled as 109.0%, 12 g water is added. The new labelling of the oleum sample is

Options

A.106.0%
B.103.0%
C.102.8%
D.105.6%

Solution

200 g of 109% oleum requires 18 g of water for complete conversion into H2SO4. On adding 12 g of water, the resulting 212 g oleum is still requiring 6 g of water. Hence, the new labelling is

$$\left( 100 + \frac{6}{212} \times 100 \right) = 102.83\%$$

Create a free account to view solution

View Solution Free
Topic: Mole Concept·Practice all Mole Concept questions

More Mole Concept Questions

Fluorine reacts with uranium hexafluoride UF6 as represented by the following equation.U(s) + 3F2(g) → UF6(g)How many fl...The explosion of a mixture consisting of one volume of a gas being studied and one volume of H2 yielded one volume water...Hydrogen cyanide, HCN, is prepared from ammonia, air and natural gas (CH4) by the following process. 2NH3(g) + 3O2(g) + ...The distribution of molecular kinetic energy at two temperatures is as shown in the following graph:Which of the followi...Twenty molecules of SO3 will weigh as much as ….. molecules of oxygen....