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The amount of K(2) Cr(2)O(7) ( eq. wt. ...

The amount of `K_(2) Cr_(2)O_(7)` ( eq. wt. 49.04) required to prepare 100 ml of its 0.05N solution is

A

2.9424 g

B

0.4904 g

C

1.4712 g

D

0.2452 g

Text Solution

Verified by Experts

The correct Answer is:
D
`W=(Nxxeq.wt.xxV(mL))/(1000)=(0.05xx49.04xx100)/(1000)=0.2452`
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