JEE Main On 8 April 2017 Question 9
Question: Excess of NaOH (aq) was added to 100 mL of $ FeCl _3 $ (aq) resulting into 2.14 g of $ Fe{{(OH)}_3}. $ The molarity of $ FeCl _3 $ (aq) is : (Given molar mass of $ Fe=56gmo{l^{-1}} $ and molar mass of $ Cl=35.5gmo{l^{-1}} $ ) [JEE Online 08-04-2017]
Options:
A) 0.3 M
B) 0.2 M
C) 0.6 M
D) 1.8 M
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \underset{\begin{smallmatrix} 100ml \\ m=? \end{smallmatrix}}{\mathop{3NaOH+Fecl _3}},\to \underset{2.14gm}{\mathop{Fe{{(CH)}_3}+NaCl}}, $ Moles of $ Fe(CH _3)=\frac{2.14}{107}=2\times {10^{-2}}mol $ moles $ FeCl _3=2\times {10^{-2}}mol $ $ M=\frac{2\times {10^{-2}}}{100}\times 1000=0.2M $
