An unknown compound is immiscible with water. It is steam distilled at `98.0^(@)C`. At `98.0^(@)C`, p and `p_(H_(2)O)^(@)` are 737 and 707 torr respectively. This distillate was 75% by weight of water. The molecular weight of the unknown will be

Options

A. `318.15 g mol^(-1)`
B. `300 g mol^(-1)`
C. `306.76 g mol^(-1)`
D. None of these

Answer

a

Explanation

Since, unknown compound is immiscible with water.
Hence, vapour pressure `prop` moles
Given, `p_("total")^(@) = 737 torr`
`p_(H_(2)O)^(@) = 707 torr`
`:. P_("unknown")^(@) = 737 - 707 = 30 torr`
`W_(H_(2)O) = 100 g`
`W_("unknown") = 75 g`
`(p_("unknown")^(@))/(p_("water")^(@)) = (n_("unknown"))/(n_(H_(2)O)) = (W_("unknown")xxm_(H_(2)O))/(W_(H_(2)O)xxm_("unknown")`
`(30)/(707) = (75.0xx18)/(100xxm_("unknown")) or m_("unknown") = 318.15 g mol^(-1)`