Question

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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Answer 1

Explanation:

In the first sample of the oxide,

wt of metal =1.80g, wt of oxygen=(3.0-1.80)g=1.2g=1.80g, wt of oxygen=(3.0-1.80)g=1.2g

wt. of metal wt.of oxygen=1.80g 1.2g31.5.:wt.

of metal wt.of oxygen=1.80g1.2g=1.5

In the second sample of the oxide, wt of metal =1.50g=1.50g wt of

oxygen =(2.50-1.50)g=1g=(2.50-1.50)g=18

wt.of metal wt of oxygen=1.50g1g=1.5.:wt.of metal wt 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 isotopes.

Answer 2

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 gm

$\tt{:\implies \dfrac{Weight \:of \:metal}{Weight \:of \:oxygen}}$

$\tt{:\implies \cancel{\dfrac{1.80}{1.5}}}$

$\bf{:\implies \underline{ \: \: \underline{ \red{ \: \: 1.5 \: \: }} \: \: }}$

Case ② :

→ Weight of metal = 1.50 gm

→ Weight of oxygen = 2.5 - 1.50

→ Weight of oxygen = 1 gm

$\tt{:\implies \dfrac{Weight \:of \:metal}{Weight \:of \:oxygen}}$

$\tt{:\implies \cancel{\dfrac{1.50}{1}}}$

$\bf{:\implies \underline{ \: \: \underline{ \red{ \: \: 1.5 \: \: }} \: \: }}$

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.

Hence Solved !