1.

8 g of HBr is added in 100 g of H_(2)The freezing point will be (K_(f) = 1.86, H=1, br = 80 )

Answer»

`-0.75^(@)C`
`0^(@) `
`-3.67 ^(@) C`
`-7.6 ^(@) C `

Solution :`Delta T_(F) = iK_(f) xx (W_(2) xx 1000)/(M_(2) xx W_(1)) = 2 xx 1.86 xx (8 xx 100)/(81 xx 100)=3.67^(@)`
`T_(f) = T^(@) - Delta T_(f) = -3.67^(@) C`


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