Q. What is the $%$ of $H_{2} O$ in $F e (CNS)_{3} .3 H_{2} O$ ?

2392 254 Some Basic Concepts of Chemistry Report Error

A

$45$

18%

B

$30$

33%

C

$19$

33%

D

$25$

15%

Solution:

In $F e (CNS)_{3} .3 H_{2} O$
% of $H_{2} O = \frac{3 \times 18}{284} \times 100 = 19%$