In an ore the only oxidisable material is Sn^(2+). This ore is titrated with a dichromate solution containing 2.5 g K_(2)Cr_(2)O_(7) in 0.50 litre. A 0.40 g of sample of the ore required 10.0 cm^(3) of the titrant to reach equivalent point. Calculate the percentage of tin in ore. (K=39.1, Cr=52, Sn=118.7)

In an ore the only oxidisable material is Sn^(2+). This ore is titrate

1.	In an ore the only oxidisable material is Sn^(2+). This ore is titrated with a dichromate solution containing 2.5 g K_(2)Cr_(2)O_(7) in 0.50 litre. A 0.40 g of sample of the ore required 10.0 cm^(3) of the titrant to reach equivalent point. Calculate the percentage of tin in ore. (K=39.1, Cr=52, Sn=118.7)
Answer» Solution :Mol. Mass of `K_(2)Cr_(2)O_(7)=2xx39.1+2xx52+7xx16` =78.2+104.0+112.0 =294.2 Eq. mass of `K_(2)Cr_(2)O_(7)=(294.2)/(6)=49.03` NORMALITY of `K_(2)Cr_(2)O_(7)` solution `=(2.5)/(49.03)xx(1000)/(500)=(5)/(49.03) N` `10 mL (5)/(49.03) N K_(2)Cr_(2)O_(7)-=10 mL (5)/(49.03)N` stannous ION Eq. mass of `Sn^(2+)=(118.7)/(2)=59.35` Amount of Sn in the sample `=(5)/(49.03)xx(59.35)/(1000)xx10` =0.0605 g Percentage of Sn in the ore `=(0.0605)/(0.40)xx100=15`