Solutions Question 156
Question: The amount of $ K_2Cr_2O_7 $ (eq. wt. 49.04) required to prepare 100 ml of its 0.05 N solution is [JIPMER 2002]
Options:
A) 2.9424 g
B) 0.4904 g
C) 1.4712 g
D) 0.2452 g
Show Answer
Answer:
Correct Answer: D
Solution:
$ W=\frac{N\times eq.wt.\times V(ml)}{1000}=\frac{0.05\times 49.04\times 100}{1000} $
$ =0.2452. $
