The molal freezing point for water is `1.86^(@)C mol^(-1)`. If 342g of cane sugar is dissolved in 1000 mL of water, the solution will freeze at

Options

A. `1.86^(@)C`
B. `-1.86^(@)C`
C. `2.42^(@)C`
D. `-2.42^(@)C`

Answer

B

Explanation

`Delta T_(f) = K_(b) xx W_(2)/(M_(2) xx W_(1) (kg))`
`Delta T_(f) = 1.86 xx 342/(342 xx 1 ) = 1.86^(@)C`
(for water d=1g ` "mol"^(-1))`
`T_(f) = T^(@)- Delta T_(f) = 0^(@)C-1.86^(@)C=-1.86^(@)C`