QUETION:−
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1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in steam gave 2.50 g of its oxide. Show that these results illustrate the law of constant proportion.
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Answers
Given :
1.80 g of a certain metal burnt in oxygen gave 3.0 gm of its oxide.
1.50 g of the same metal heated in steam gave 2.50 gm of its oxide.
To Find :
Show that these results illustrate the law of constant proportion.
Required Solution :
Case ① :
→ Weight of metal = 1.80 gm
→ Weight of oxygen = (3.0 - 1.80)
→ Weight of oxygen = 1.2
:⟹ Weightofoxygen
Weightofmetal
⟹ 1.5
1.80
1.5 ⟹
1.5
Case ② :
→ Weight of metal = 1.50 gm
→ Weight of oxygen = 2.5 - 1.50
→ Weight of oxygen = 1 gm
\tt{:\implies \dfrac{Weight \:of \:metal}{Weight \:of \:oxygen}}:⟹
Weightofoxygen
Weightofmetal
\tt{:\implies \cancel{\dfrac{1.50}{1}}}:⟹
1
1.50
\bf{:\implies \underline{ \: \: \underline{ \red{ \: \: 1.5 \: \: }} \: \: }}:⟹
1.5
In both Case ① and Case ② Proportional of metal are constant, In Case ① 1.5 and in Case ② also 1.5 that means this follow law of constant proportion.
Answer:
Explanation
In the first sample of the oxide,
wt. of metal=1.80 g, wt. of oxygen =(3.0-1.80)g =1.2 g
∴wt. of metalwt. of oxygen=1.80g1.2g=1.5
In the second sample of the oxide,
wt of metal =1.50 g, wt. of oxygen=(2.50-1.50)g =1 g
∴wt. of metalwt. of oxygen=1.50g1g=1.5
Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen are fixed. Hence, the results follows the law of constant proportion.
Note : This law is not applicable in case of isotopes.