The boiling point of water (`100^@ C`) becomes, `100.52^@ C` if 3 g of non-volatile solute is dissolved is 200 ml of it: Calculate the molecular weight of the solid. (`K_b` = 0.6 K/m)

Explanation

`Delta T_b = 100.52 -100=0.52 ^@ C K`
` W_B =3g`
`W_A =200 ml`
` M_B=?`
` K_b = 0.6 K//m`
` Delta T_b =K_b xx(W_b xx 1000 )/( M_B W_A)`
` 0.52 =( 0.6 xx 3xx 1000)/( 200xx 0.52 )`
`=( 90 )/( 0.52)`
`=173.07//Mol .`