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. $



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