Question

1.80 gram of a certain metal burnt in oxygen and gave 3.0 gram of its oxide. 1.50 gram of the same element heated in the same gave 2.50 gram of its oxide. Show that these result illustrate the law of constant proportion.
Need your help...​

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 metalwt.of oxygen=1.80g1.2g=1.5∴wt. of metalwt.of oxygen=1.80g1.2g=1.5 

In the second sample of the oxide, 

wt of metal =1.50g=1.50g wt of oxgen =(2.50−1.50)g=1g=(2.50-1.50)g=1g 

∴wt.of metalwt of oxygen=1.50g1g=1.5∴wt.of metalwt of oxygen=1.50g1g=1.5 

Thus in both smaples of the oxide the proportions of the weights of the metal and oxgen are fixed. Hence, the results follows the law of constant proportion 

Note This law is not applicable in isotopes .

Answer 2

Given:

In sample 1

Mass of metal, $\[{m_{M1}} = 1.80\,g\]$

Mass of metal oxide, $\[{m_{MO1}} = 3.0\,g\]$

In sample 2

Mass of metal, $\[{m_{M2}} = 1.50\,g\]$

Mass of metal oxide, $\[{m_{MO2}} = 2.50\,g\]$

To Prove: The results illustrate the law of constant proportion

Solution:

In the first sample,

Mass of oxygen can be calculated as:

$\[\begin{gathered} {m_{O1}} = 3.0\,g - 1.80\,g \hfill \\ {m_{O1}} = 1.2\,g \hfill \\ \end{gathered} \]$

Therefore, the ratio of the mass of metal to the mass of oxygen is given as:

$\[\begin{gathered} \frac{{{\text{mass}}\,{\text{of}}\,{\text{metal}}}}{{{\text{mass}}\,{\text{of}}\,{\text{oxygen}}}}{\text{ = }}\frac{{1.8}}{{1.2}} \hfill \\ \frac{{{\text{mass}}\,{\text{of}}\,{\text{metal}}}}{{{\text{mass}}\,{\text{of}}\,{\text{oxygen}}}}{\text{ = 1}}{\text{.5}} \hfill \\ \end{gathered} \]$

In the second sample,

Mass of oxygen can be calculated as:

$\[\begin{gathered} {m_{O2}} = 2.5\,g - 1.50\,g \hfill \\ {m_{O2}} = 1.0\,g \hfill \\ \end{gathered} \]$

Therefore, the ratio of the mass of metal to the mass of oxygen is given as:

$\[\begin{gathered} \frac{{{\text{mass}}\,{\text{of}}\,{\text{metal}}}}{{{\text{mass}}\,{\text{of}}\,{\text{oxygen}}}}{\text{ = }}\frac{{1.5}}{{1.0}} \hfill \\ \frac{{{\text{mass}}\,{\text{of}}\,{\text{metal}}}}{{{\text{mass}}\,{\text{of}}\,{\text{oxygen}}}}{\text{ = 1}}{\text{.5}} \hfill \\ \end{gathered} \]$

As in both the samples, the proportions of mass of metal and mass of oxygen are the same, therefore, the results illustrate the law of constant proportions.

Hence, these results illustrate the law of constant proportion.

#SPJ2