The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. `K_b`, for benzene is `2.53 K kg mol^(-1)`.

Options

A. `58 g mol^(-1)`
B. `85 g mol^(-1)`
C. `93 g mol^(-1)`
D. `108 g mol^(-1)`

Answer

A

Explanation

`DeltaT_b = 354.11 - 353.23 = 0.88 K`
`w_A = 1.80 g, m_A = ?, w_B = 90`
`DeltaT_b= K_b m= K_bxx (w_A//m_A)/(W_B)xx1000`
`0.88= 2.35 xx (1.80//m_A)/90 xx 1000 implies m_A = (2.53 xx 1.80 xx 1000)/(90 xx 0.88)`
`= 57.5 ~~ 58 g mol^(-1)`