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Before dissolution, the substance is usually a solid and forms crystals. Its physical state is solid. After dissolution, the substance is not visible any more : it is not a solid any more. Its physical state has changed. Before hydration, the substance is usually solid. After hydration, the color and the volume may have changed, but the physical state has ...


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You make at least one mistake in your original calculation: the dilution factor should be $\times 100$ not $\times 1000$, which importantly reduces the concentration in the original sample to $\pu{0.2 \mu g / \mu L}$ or $\pu{200 ng / \mu L}$ RNA and makes it rather trivial to see what volume is required to obtain $\pu{100 ng}$, clearly this must be $\frac12 ...


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To derive it, you have to know how an equilibrium depends on concentration and on temperature. At the freezing point, pure liquid and the solid are at equilibrium. If you lower to concentration of the liquid by adding solute, you disturb the equilibrium. If you lower the temperature by the right amount, you are back at equilibrium. The relationship you get ...


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Let's start with the mixture $x_A$, which is made of 85% A + 15 % B. When heating this liquid, the first bubble vaporized appeared at a temperature just a little above $T'_A$, as given by the horizontal line. And the composition of the vapor is nearly pure $A$, with $1$% $B$. As a consequence, the liquid is loosing some $A$. its volume decreases, but as it ...


1

If you have two component mixture of A and B, most important reason for volume of mixing being zero is that when solution is ideal heterogeneous interactions (A - B) are the same as homogeneous (A - A and B - B). If interactions are the same than volume doesn't change when you mix components. Other reasons which contribute to non - ideality of mixing is size ...


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Your solution is probably made of aluminum sulfate $\ce{Al2(SO4)3}$, which contains aluminum ions $\ce{Al^{3+}}$, due to the reaction $$\ce{Al2(SO4)3 -> 2 Al^{3+} + 3 SO4^{2-}}$$ And the $\ce{Al^{3+}}$ ions react slowly with water according to : $$\ce{Al^{3+} + 3 H2O <=> Al(OH)3(s) + 3 H+}$$ This equation is an equilibrium. You should see that there ...


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At very dilute concentrations, the solvating power of water overcomes the natural tendency of surfactant molecules to agglomerate into a separate phase. The shear numerical excess of water molecules just about totally dissociates the bulk surfactant and dissolves it as separate molecules. So a mixture of surfactant and water can be a true solution - clear ...


3

The point of the question is to test your understanding i. of Raoult's law (that increasing the solute concentration decreases the vapor pressure and that the identity of the solute is not important in an ideal solution; the mole fraction of solute determines $\Delta p$: $\Delta p = p-p^\circ = -\chi_\textrm{solute}p^\circ$ ii. that vapor pressure increases ...


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In his answer Dario Miric has explained the basis of azeotropes clearly, this is a small addition to his answer. Raoult determined experimentally that the partial vapour pressure over some liquid mixtures is directly proportional to the mole fraction $x$ of each component, and thus the total pressure is $$P=x_1p_1^\text{o}+x_2p_2^\text{o}=x_1p_1^\text{o}+(1-...


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To understand why non - ideal mixtures form azeotropes, we need to understand what is ideal mixture. If you have a liquid mixture of two components A and B, ideal mixture is a mixture in which heterogeneous interactions (A - B) are the same as the average homogeneous (A - A and B - B). This means when you mix components A and B, they don't "feel" ...


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It is not a messed up cycle. Let's consider what happens with an example. If you heat up a mixture containing $10$% ethanol + $90$% water, the mixture boils at $92$°C, producing a vapor containing $50$% ethanol + $50$% water. With this operation the liquid loses more ethanol than water. So its concentration in ethanol decreases, and it is necessary to heat ...


2

The starting point to solve the problem is Raoult's law. For each component we can write that the vapour pressure (related by Dalton's law to the total pressure) is equal to the product of the mole fraction in the solution and the vapour pressure of the pure component: $$y_iP_S = \chi _i P^{\circ} _i$$ You can exploit the fact that the molar amounts in the ...


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