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Water molecules in gas phase collide and are absorbed by both solutions at the same rate, assuming the same humidity over the solutions. OTOH, as sugar solution has lower molar fraction of water than water itself, its rate of water evaporation is lower than pure water evaporation. Therefore, there is the net transfer of water molecules from the water beaker ...


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The water goes to the sugar beaker spontaneously, be careful for this word, in that you don't make any effort to move it there except only connecting the two beakers. Any spontaneous process have a negative free energy. The negative free energy is the result of comparing the energy state of the system before the process (water only beaker) and the energy ...


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Imagine the converse situation: A sealed chamber has an open beaker of the volatile solute. After some time, equilibrium vapor is established. Now an open beaker of the non-volatile solvent is introduced into the chamber. Some molecules of the solvent in the chamber's "atmosphere" will dissolve into the open beaker of solvent, and that process ...


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For solvent at temperature $T$, surface area $A$ and having the molar fraction $x$, the rate of molecules leaving liquid is: $$\frac {\mathrm{d}n}{\mathrm{d}t}=-C_1 \cdot A \cdot x \cdot f(T)$$ Lower molar fraction means lower surface density and lower evaporation rate per surface area. Factor $C_1$ includes pure solvent molecular surface density and $f(T)$ ...


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If we add non volatile solute to solvent for example water, we decrease the tendency of water to evaporate in gas phase because solute particle obstructs the evaporation of water in gas phase therefore vapour pressure decrease when we add non volatile solute to solvent like water. This image is from ncert:


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It is valid for both volatile and nonvolatile solutes, as it refers to the partial vapour pressure, not total vapour pressure. Saying that, note that it applies only on mixtures with ideal behaviour, as there are many more or less significant positive and negative deviations from the law. These deviations relate to preference of intermolecular bonding to ...


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You must understand... in an experiment, the sigmoidal shape will never have infinite slope... This would be a discontinuity... the sigmoidal shape (rightmost) and the "curve" (middle image) are essentially representing identical phenomena... there is a transition from one pure component to another. The only difference is how abrupt the change is......


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Create a vacuum enduring apparatus, equipped with a heated flask with the solution, with constrained capillary air inlet immersed into the liquid ( provides boiling centers to avoit overheating ) and pressure sensor, attached to a vacuum source. Combination of applied heating and vacuum, there will be established "laboratory grade equilibrium", ...


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I'd like to specifically commment on this: According to the NCERT for Class XII, Part I, [pg. 46, para 3][1], In a pure liquid the entire surface is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give a solution [Fig. 2.4.(b)], the vapor pressure of the solution is solely from the solvent ...


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Vapor pressure is the pressure exerted by the vapors when in dynamic equilibrium with its condensed phase at a given temperature. It is the maximum value of partial pressure at a given temperature. So if pressure is increased after this value then some vapors will condense and if pressure is decreased then more liquid will vaporise until the vapor pressure ...


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You may want to read the Wikipedia article on TMAH. While it does decompose, it may not produce the methanol you are expecting. Also, heat is also involved in this process, so you may need to heat and, in the process, vaporize much of the water to get this decomposition to go. TMAH, in common with many other TMA salts containing simple anions, decomposes ...


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