When 2 g of a non-volatile solute was dissolved in 90 g of benzene the boiling point of benzene is raised by 0.88 K. Which of the following may be the solute? `(K_b " for benzene " = 2.53 K kg mol^(-1)` )

Options

A. `CO(NH_2)_2`
B. `C_6H_(12)O_6`
C. NaCl
D. None of these.

Answer

D

Explanation

`w_A= 2 g, w_B = 90 g, M_B = 78`
`DeltaT_b = 0.88, K_b = 2.53 K kg mol^(-1)`
`DeltaT_b= K_bxx m`
`0.88 = 2.53xx(2//m_A)/90 xx1000 implies 0.88 = (2.53xx2)/(90 xxm_A)xx1000`
`m_A = (2.53 xx 2 xx 1000)/(90xx0.88) = 63.8 ~~ 64`
Molecular Mass of `CO(NH_2)_2 implies 60`
Molecular Mass of `C_6H_(12)O_6 implies 180`
Molecular Mass of `NaCl implies 58.5`