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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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Answer» `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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