The molal freezing point for water is 1....

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

`1.86^(@)C`

`-1.86^(@)C`

`2.42^(@)C`

`-2.42^(@)C`

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The correct Answer is:

`m = (1000K_(f)w)/(DeltaT_(f)W)`
`DeltaT_(f) = (1000K_(f)w)/(mW) = (1000x1.86xx342)/(342xx1000) = 1.86^(@)C`
`:.` The solution will freezes at = `0 - 1.86^(@)C = -1.86^(@)C`

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The molal freezing point constant of water is 1.86 K m^(-1) . If 342 g of cane sugar (C_(12)H_(22)O_(11)) is dissolved in 1000 g of water, the solution will freeze at

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