Question

0.1 g of metal combines with 46.6 mL of oxygen at STP. The equivalent metal is:
A. 12
B. 24
C. 6
D. 36

Answer 1

Hint: From Avogadro’s law it is known that equal volumes of all gases contain equal numbers of molecules under similar conditions. Now since one mole molecule of all gases contains the same number of molecules therefore they occupy the same volume under similar conditions of temperature and pressure.

Complete step by step answer:
The volume occupied by one mole molecules of a gaseous substance is called molar volume. In the question above the metal combines with oxygen under standard conditions of temperature and pressure we know one mole molecules of all gases occupy 22.4 litres at 273 K and 760 mm pressure (S.T.P).
Hence we can conclude that the molar volume of all gases at STP is 22.4 L or 22400 ml. Hence one mole of oxygen will have the equal weight in grams as the molar mass of oxygen, therefore we can conclude(1 mole of${O_2}$ =32 g of ${O_2}$₎
We know gram equivalent is fixed and for oxygen it is,
1 g equivalent = 8 g of ${O_2}$ (molar mass of oxygen is 32 g)
Therefore gram equivalent in 1 mole of oxygen is 32/8 g equivalent= 4 g equivalent. Now at STP since all molecules occupy the same volume that is 22400 mL then 4 g equivalent of oxygen will also occupy the same.
22400 mL of ${O_2}$ = 4 eq of O
46.6 mL of ${O_2}$ = $\dfrac{4}{{22400}} \times 46.6 = 0.00832\;eq$
Since 0.1 g of metal combined with oxygen equivalent to metal will be equal to the equivalent of oxygen.
The equivalent weight of the metal is given by ${\dfrac{Mass \;of \;metal \;given}{Equivalent\; of\; oxygen}}$
Equivalent of metal =$\dfrac{{0.1}}{{0.00832}} = 12.0$ eq.

Hence the correct option is option A.

Note: Calculation needs to be done carefully. Equivalent weight is the amount of weight or amu that combines with the amount of substance in grams.