The molal b.pt constant for water is `0.513^(@)C "kg mol"^(-1)`. When 0.1 mole of sugar is dissolved in 200g of water, the solution boils under a pressure of 1 atm at :
A. `100.513^(@)C`
B. `100.0513^(@) C`
C. `100.256^(@)C`
D. `101.025^(@)C`

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1 Answers

`Delta T_(b) = K_(b) xx m 0.51 xx (0.1/200 xx 1000)=0.2565^(@)C`
`T_(b) = T^(@) + Delta T_(b) = 100 + 0.2565^(@) C=100.2565^(@) C`

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