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Hint: As the metal is x grams and oxygen is y grams, the amount of metal in the given compound will be (x-y) grams. As the atomic number of oxygen is 8, the formula for equivalent weight of the changes accordingly.
Complete answer:
Here, given metal oxide of x gm, the mass of oxygen in it is y gm, thus the mass of metal becomes (x-y) gm.
Mass of metal = Mass of metal oxide – Mass of oxygen
Mass of metal = (x-y) gm
Now, the atomic number of oxygen is 8, therefore the formula for equivalent weight of metal in metal oxide is \[\dfrac{8(x-y)}{y}\] .
Thus, the equivalent weight of metal is equal to \[\dfrac{8(x-y)}{y}\] .
Equivalent weight of any substance is the mass of one equivalent, which is the mass of a substance which will combine with a fixed quantity of another substance.
Thus, here the metal will combine with 8 gm of oxygen to give the compound which is metal oxide.
Atomic weight does not have any unit, but equivalent weight has the unit of mass as gram.
This similar way can also be used to find equivalent weight of metals in different metal oxides.
Note:
The atomic number of oxygen needs to be remembered as it is important to find equivalent weight of any metal in metal oxide. This method can be used to find equivalent weights of metals in metal oxides.
Complete answer:
Here, given metal oxide of x gm, the mass of oxygen in it is y gm, thus the mass of metal becomes (x-y) gm.
Mass of metal = Mass of metal oxide – Mass of oxygen
Mass of metal = (x-y) gm
Now, the atomic number of oxygen is 8, therefore the formula for equivalent weight of metal in metal oxide is \[\dfrac{8(x-y)}{y}\] .
Thus, the equivalent weight of metal is equal to \[\dfrac{8(x-y)}{y}\] .
Equivalent weight of any substance is the mass of one equivalent, which is the mass of a substance which will combine with a fixed quantity of another substance.
Thus, here the metal will combine with 8 gm of oxygen to give the compound which is metal oxide.
Atomic weight does not have any unit, but equivalent weight has the unit of mass as gram.
This similar way can also be used to find equivalent weight of metals in different metal oxides.
Note:
The atomic number of oxygen needs to be remembered as it is important to find equivalent weight of any metal in metal oxide. This method can be used to find equivalent weights of metals in metal oxides.
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