
Given, ${{\text{H}}_{\text{2}}}\text{S}$ contains 5.88% hydrogen, ${{\text{H}}_{\text{2}}}\text{O}$ contains 11.11% hydrogen while $\text{S}{{\text{O}}_{\text{2}}}$ contains 50% sulphur. These figures illustrate the law of:
(A) conservation of mass
(B) constant proportions
(C) multiple proportions
(D) reciprocal proportions
Answer
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Hint: Find out the molecular weight of each compound. Recollect the basic laws of chemical combination.
Complete answer:
-According to the law of definite proportions, “Any pure compound always contains the same elements in a definite proportion by weight irrespective of its source or method of preparation.” This law was given by French chemist, Joseph Proust. It is also known as the law of constant proportions.
-The molecular weight of ${{\text{H}}_{\text{2}}}\text{S}$ is equal to 34g/mol. There are two hydrogen atoms, so the mass of hydrogen in ${{\text{H}}_{\text{2}}}\text{S}$ is 2g. So, the amount of hydrogen present in ${{\text{H}}_{\text{2}}}\text{S}$ is $\dfrac{2}{34}\times 100=5.88%$
-Similarly, let's calculate hydrogen in water. The molecular weight of ${{\text{H}}_{\text{2}}}\text{O}$ is 18g. There are two hydrogen atoms, so the mass of hydrogen in ${{\text{H}}_{\text{2}}}\text{O}$ is 2g. So, the amount of hydrogen present in ${{\text{H}}_{\text{2}}}\text{O}$ is $\dfrac{2}{18}\times 100=11.11%$
-In the same manner, let us calculate sulphur in sulphur dioxide,$\text{S}{{\text{O}}_{\text{2}}}$. The molecular weight of $\text{S}{{\text{O}}_{\text{2}}}$ is 64g. There is one sulphur atom present in $\text{S}{{\text{O}}_{\text{2}}}$. So, the mass of sulphur is 32g. Therefore, the amount of sulphur present in $\text{S}{{\text{O}}_{\text{2}}}$ is $\dfrac{32}{64}\times 100=50%$
-Thus, we can conclude that the given figures illustrate the law of constant proportions.
So, the correct option is (B) constant proportions.
Note:
Don’t get confused with other laws of combination like law of conservation of mass and law of multiple proportions. The law of multiple proportions talks about finding the ratios of masses of two or more elements of the compound in smaller numbers instead of the percentage proportion of each element in the compound.
Complete answer:
-According to the law of definite proportions, “Any pure compound always contains the same elements in a definite proportion by weight irrespective of its source or method of preparation.” This law was given by French chemist, Joseph Proust. It is also known as the law of constant proportions.
-The molecular weight of ${{\text{H}}_{\text{2}}}\text{S}$ is equal to 34g/mol. There are two hydrogen atoms, so the mass of hydrogen in ${{\text{H}}_{\text{2}}}\text{S}$ is 2g. So, the amount of hydrogen present in ${{\text{H}}_{\text{2}}}\text{S}$ is $\dfrac{2}{34}\times 100=5.88%$
-Similarly, let's calculate hydrogen in water. The molecular weight of ${{\text{H}}_{\text{2}}}\text{O}$ is 18g. There are two hydrogen atoms, so the mass of hydrogen in ${{\text{H}}_{\text{2}}}\text{O}$ is 2g. So, the amount of hydrogen present in ${{\text{H}}_{\text{2}}}\text{O}$ is $\dfrac{2}{18}\times 100=11.11%$
-In the same manner, let us calculate sulphur in sulphur dioxide,$\text{S}{{\text{O}}_{\text{2}}}$. The molecular weight of $\text{S}{{\text{O}}_{\text{2}}}$ is 64g. There is one sulphur atom present in $\text{S}{{\text{O}}_{\text{2}}}$. So, the mass of sulphur is 32g. Therefore, the amount of sulphur present in $\text{S}{{\text{O}}_{\text{2}}}$ is $\dfrac{32}{64}\times 100=50%$
-Thus, we can conclude that the given figures illustrate the law of constant proportions.
So, the correct option is (B) constant proportions.
Note:
Don’t get confused with other laws of combination like law of conservation of mass and law of multiple proportions. The law of multiple proportions talks about finding the ratios of masses of two or more elements of the compound in smaller numbers instead of the percentage proportion of each element in the compound.
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