Calculate the vapor pressure of 0.5 molal AQ solution of sucrose at 100 degree Celsius

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Solution

$\frac{P^{0} - P}{P^{0}} = \frac{W_{2} \times M_{1}}{M_{2} \times W_{1}} - - - - - - 1 B u t m o l a l i t y (m) = \frac{W_{2} \times 1000}{M_{2} \times W_{1}} \frac{W_{2}}{M_{2} \times W_{1}} = \frac{m}{1000} S u b s t i t u t e t h e a b o v e i n e q u a t i o n 1 \frac{P^{0} - P}{P^{0}} = \frac{m \times M_{1}}{1000} v p o f w a t e r a t 100^{0} C i s 1 b a r \frac{1 - P}{1} = \frac{0.5 \times 18}{1000} 1 - P = 0.009 P = 1 - 0.009 = 0.991 b a r$

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Calculate the vapor pressure of 0.5 molal AQ solution of sucrose at 100 degree Celsius

P0 - PP0 = W2× M1M2×W1 ------1But molality (m) = W2× 1000M2×W1 W2M2×W1 = m1000Substitute the above in equation 1P0 - PP0 = m× M11000 vp of water at 100 0C is 1 bar1 - P1 = 0.5× 1810001-P = 0.009P = 1 - 0.009 = 0.991 bar