A solution containing 28 g of phosphorus in 315 g `CS_(2)(b.p. 46.3^(@)C`) boils at `47.98^(@)C` . If `K_(b)` for `CS_(2)` is `2.34` K kg `mol^(-1)` . The formula of phosphorus is (at , mass of P = 31).

Answer

4

Explanation

`DeltaT = (1000 xx K_(b)^(.) xx w)/(m xx W)`
1.68 `=(1000 xx 2.34 xx 28)/(m xx 315)`
`therefore m_(exp) = 123.80`
`m_(N)/m_(exp) =1- alpha + alpha/n`
`therefore alpha=1, therefore m_(N)/m_(exp) =1/n (m_(N) " of " P=31)`
`therefore 31/(123.80) = 1/n`
`therefore n=4`