Correct Answer - 2
`2Na^(+)(aq)+2I^(-)(aq)+HgI_(2)(g)rarrNa^(+)(aq)+HgI_(4)^(2-)(aq)`
The number of mole particle decrease from `4(2Na^(+)+2I^(-))` to to `3(2Na^(+)+HgI_(4)^(2-))`. Hence, the colligative property will decrease ot, the vapour pressure will increase to a constant value until NaI is completely consumed.