A solid mixture of 5 g of lead nitrate and sodium nitrate was heated below `600^(@)C` until mass of residue was constant. If the loss in mass is 28%, find the mass of lead nitrate (a) and sodium nitrate (b) in mixture.

Options

A. `a = 4. 21 g, b = 0.12 g`
B. `a = 5 g, b = 2. 16 g`
C. `a = 3.32 g, b = 1. 68 g`
D. `a = 1.23 g , b = 4. 20 g`

Answer

C

Explanation

`Pb underset( ag) ((NO _(3))) _(2) to Pb O + 2 NO _(2) uarr + (1)/(2) O _(2) uarr , underset(bg) ( NaNO _(3)) to Na NO _(2) + (I)/(2) O _(2) uarr`
`therefore a + b = 5 " "...(1)`
The loss in mass for 5 g mixture `= 5 xx (28)/(100) = 1. 4 g .` The residues left `= 5 - 1.4 =3.6 g `
The resisdue contains `PbO + N aNO _(2) therefore a g Pb ( NO _(3)) _(2)` gives `= (223 xx a)/( 331) g PbO`
`therefore b g NaNO _(3) ` gives `(69 xx b)/(83) g Na NO _(2) .` Similarly, 85 g `NaNO _(3)` gives `= 69 g NaNO _(2)`
`therefore (223 xxa)/( 331) + ( 69 xx b)/( 85) = 3. 6 " "...(2)`
Solving Eqs. (1) and (2) `implies a = 3.32 g , b = 1. 68 g `