Question

0.12gm of magnesium is treated with an acid which gives 0.60gm of anhydrous magnesium salt then equivalent weight of an acid is

Answer 1

Solution.

n(Mg) = 0.005 mole

n(Mg) = n(salt) = 0.005 mole

$$M(salt) = \frac{m}{n}$$

M(salt) = 121.55

$$M(acid) = M(salt)-M(Mg)$$

M(acid) = 97.24

$$M^{e} (acid) = \frac{97.24}{2}$$ $$M^e(acid) = 48.62$$

Answer:

$$M^e(acid) = 48.62$$