The vapour pressure of pure benzene ` (C_6H_6) ` at a given temperature is 640 mm Hg . 2.2 g of non -volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg . Calcultaed the molar mass of the solute?

Explanation

` P_(C_6H_6) ^(@) = 640 mm Hg `
`W_2 = 2.2 g ` (non volatile solute )
` W_1 = 40 g ` ( benzene)
` P_( " solution " ) = 600 mm Hg `
` M_2 =? `
` (P^(@) -P) /( P^(@) ) =x_2 `
` ( 640- 600)/( 640) =( n_2 )/( n_1+n_2) ,`
` ( 40 )/( 640)=( n_2) /( n_1) `
` 0.0625=(W_2 xx M_1)/( M_2 xx W_1 ) `
` M_2 =( 2.2xx 78)/( 0.0625xx 40)= 68.64 mol ^(-1) `