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

2.9424 g

`0.4904g `

`1.4712g`

`0.245g`

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The correct Answer is:

1 Lit. 1N `K_(2)Cr_(2)O_(7)` required 49.04 gms of `K_(2)Cr_(2)O_(7)`.
1 Lit. 0.05 N `K_(2)Cr_(2)O(7)` required 2.45 gms of `K_(2)Cr_(2)O_(7)`
`:.` 100ml 0.05 N `K_(2)Cr_(2)O_(7)` required 0.245 gms of `K_(2)Cr_(2)O_(7)`.

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