In an ore, the only oxidizable material is `Sn^(2+)`. This ore is titrated with a dichromate solution containing `2.5g` of `K_(2)Cr_(2)O_(7)` in `0.5 "litre"`. A `0.40 g` sample of the ore required `10.0 cm^(3)` of titrant to reach equivalence point. Calculate the percentage of tin in ore.

Explanation

Mol. mass of `K_(2)Cr_(2)O_(7) = 2 xx 39.1 + 2 xx 52 + 7 xx 16 `
`= 78.2 + 104.0 + 112.0`
`= 294.2`
Eq. mass of `K_(2)Cr_(2)O_(7) = ( 294)/( 6) = 49.03`
Normality of `K_(2)Cr_(2)O_(7)` solution
`= ( 2.5)/( 49.03) xx ( 1000)/( 500) = ( 5)/( 49.03) N`
10mL `( 5)/( 49.03) N K_(2)Cr_(2)O_(7)`
`= 10 mL ( 5)/( 49.03)N` stannouse ion
Eq. mass of `Sn^(2+) = (118.7)/( 2) = 59.35`
Amount of Sn in the sample `= ( 5)/( 49.03) xx ( 59.35)/( 1000) xx 10 = 0.0605g`
Percentage of Sn in the ore `= ( 0 .0605)/( 0.40) xx 100 = 51`