Question

Tin forms two chlorides, compound A and compound B. Mass percentage of tin in these chlorides are 62.63% and 45.6% respectively. Select the incorrect statement. Given data: Atomic mass of Sn is 119 amu and Cl is 35.5 amu.
A)Given data illustrates law of multiple proportion

B) Mole ratio of chloride in compound A to compound B is 1:2

C) Mole ratio of tin in compound A to compound B is 1:1

D) Equivalent weight of tin is same in both​

Answer 1

Answer:

D) Equivalent weight of tin is same in both

Explanation:

As n factors of compounds A & B are different, their equivalent weights would also be different.

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Answer 2

From the given options, Option D is the incorrect statement.

• Tin belongs to group IVA in the periodic table, The possible values of Tin are +2 and +4.
• Hence when Tin reacts with Chlorine, the possible compounds are SnCl2 and SnCl4.

Percentage of Tin in SnCl2 = ((Weight of tin)/(Total weight) )*100.

=> Percentage of Tin in SnCl2 = $\frac{119}{190} * 100$,

=> Percentage of Tin in SnCl2 = 62.63.

Percentage of Tin in SnCl4 = ((Weight of tin)/(Total weight) )*100.

=> Percentage of Tin in SnCl4 = $\frac{119}{261} * 100$,

=> Percentage of Tin in SnCl4 = 45.6.

• From the above calculations, we can conclude that Compound A is SnCl2 and Compound B is SnCl4.

• SnCl2 and SnCl4 illustrate the law o proportion. (Law of proportion states that when two elements combine in more than one way with a different chemical formula, the law of multiple proportions can be defined).
• The mole ratio of Chlorine in Compound A to Compound B is 1:2. (Compound A Contains 2 moles of Chlorine, Compound B contains 4 moles of chlorine).
• The mole ratio of tin is the same in both cases. ( SnCl2 and SnCl4 has one mole of tin in each compound)

• The Equivalent weight of Tin is not the same in both cases. ( For Compound A the Equivalent weight is (weight/2), For Compound B the equivalent weight is (weight/4). Hence the equivalent weights are not the same.

Therefore Option D is the incorrect statement.