Answer to Question #91317 in Physical Chemistry for Rajib Dhakal

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:

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