Question

Equivalent wt. of a metal is 2.5 times higher than oxygen. The ratio of weight of metal to the oxide is:
A. 0.4
B. 1.4
C. 2.4
D. 3.4

Answer 1

Hint: To solve this question, first we need to calculate the Equivalent weight of metal by using information given in the question and then after find Molecular weight of metal in terms of valence factor and also find Molecular weight of metal oxide in terms of n. After getting these two values we simply take the ratio of these two.

Formula used- As we know that eq. wt.= $\dfrac{{{\text{mol}}{\text{. wt}}{\text{. (}}{{\text{M}}_{\text{m}}}{\text{)}}}}{{{\text{valence factor (n)}}}}$

Complete step-by-step answer:
Given that the equivalent weight of a metal is 2.5 times higher than oxygen.
Equivalent weight of oxygen = 8
And we need to calculate the ratio of weight of metal to the oxide-
As we know,
∴ Equivalent weight of metal = equivalent wt. of oxygen + (2.5 × equivalent weight of oxygen)
                                                     = 8+2.5 × 8 = 28
As we know that eq. wt.= $\dfrac{{{\text{mol}}{\text{. wt}}{\text{. (}}{{\text{M}}_{\text{m}}}{\text{)}}}}{{{\text{valence factor (n)}}}}$
∴ Molecular weight of metal $({M_m})$= 28n
Formula of metal oxide will be- ${M_2}{O_n}$
Molecular weight of metal oxide = 2 × ${M_m}$+ n × ${M_m}$= 56 n + 16 n=72n
Ratio of weight of metal to its oxide = $\dfrac{{{\text{wt}}{\text{.of metal}}}}{{{\text{wt}}{\text{. of metal oxide}}}}$
= $\dfrac{{28n}}{{72n}}$= 0.389 $ \approx $0.4
Therefore, the ratio of weight of metal to the oxide is 0.4.
Hence the required answer is (A).

Note- Equivalent weight (E) unlike molecular weight is proportional mass of chemical entities which combine or displace other chemical entities. Or Equivalent weight described as the mass of a compound, elements or ion which combines or displaces 1 part of hydrogen or 8 parts of oxygen or 35.5 parts of chlorine by mass.