Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea `(NH_2CONH_2)` is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Text Solution
Verified by Experts
The data provided is as given below : `p^0 = 23.8 mm` `w_2 = 50 g , M_2 ("urea") = 60 g "mol"^(-1)` ` w_1 = 850 g, M_1 ("water") = 18 g "mol"^(-1)` Applying Raoult.s law, `(p^0 - p_x)/(p^0) = (n_2)/(n_1+ n_2) = (w_2 // M_2)/(w_1 // M_1 + w_2 // M_2)` Substituting the values, we get `(p^0 - p_x)/(p^0) = (50 // 60)/(850//18 + 50//60) = (0.80)/(47.22 + 0.83) = 0.017` Thus, relative lowering of vapour pressure = 0.017 Vapour pressure of water for this solution means `P_s` Substituting `p^0` = 23.8 mm, we get `(23.8 - p_0)/(p_s) = 0.017 " or " 23.8 -p_s = 0017 p_s` `1.017 p_s = 23.8 " or " p_s = 23.4 mm` Thus, vapour pressure of water in the solution = 23.4 mm Hg.
U-LIKE SERIES|Exercise SELF ASSESSMENT TEST ( SECTION A)|7 Videos
Similar Questions
Explore conceptually related problems
Vapour pressure of pure water at 298 K is 23.8 mm Hg . 50g of urea (NH_(2)CONH_(2)) is dissolved in 850g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
3g urea is dissolved in 45g of water. The relative lowering of vapour pressure is :
The vapour pressure of water at 23^(@)C is 19.8 mm of Hg 0.1 mol of glucose is dissolved in 178.2g of water. What is the vapour pressure (in mm Hg) of the resultant solution?
6g of urea is dissolved in 90g of water. The relative lowering of vapour pressure is equal to
A solution of non-volite solute in water freezes at -0.80^(@)C .The vapour pressure of pure water at 298 K is 23.51 mm Hg .and K_(f) for water is 1.86 degree per molal. Calculate the vapour pressure of this solution at 298 K .
The vapour pressure of water at 23^(@)C is 19.8 mm. 0.1 mole glucose is dissolved in 178.2g of water. What is the vapour pressure (in mm) of the resultant solution ?
U-LIKE SERIES-SOLUTION-SELF ASSESSMENT TEST (SECTION D)
