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# Tutor profile: Anu M.

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Anu M.
BioChemistry Tutor with 2 years of experience
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## Questions

### Subject:Physical Chemistry

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Question:

How can molality be used instead of molarity in constructing rate laws? Can mole fractions be used?

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Anu M.

The number of moles of solute per kilogram of solvent is called the molal concentration or the molality (abbreviated m), of a solution Molality, m= Moles of solute/kg of solvent Molarity or the molar concentration (abbreviated M), of a solution is the number of moles of solute per liter of solvent Molarity, M = moles of solute/ Liters of solution Consider a chemical reaction with an equation of the following form: A+ B ------> Products Its rate of reaction can be expressed as follows: Rate= k[A]^m.[B]^n where [A] and [B] are respective concentrations, the values of the exponents n and m are found by experiment and k is rate constant. Molality and molarity are concentration terms, independent of composition amount of substances other than that of interest. So we can use molality instead of molarity in constructing rate laws The mole fraction is defined the ratio of the number of moles of a given component to the total number of moles of all components XA = nA/ nA+nB+.....nZ where XA is the mole fraction of component A, and nA, nB, . . . , nZ are the numbers of moles of each component, A, B, . . . , Z, respectively. Since mole fractions are inter dependent on each other, change in one component inversely affects the mole fraction of other components. The concentration terms in the rate law needs to be independent of each other. So mole fractions cannot be used in constructing rate laws

### Subject:Organic Chemistry

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Question:

A certain organic substance is soluble in solvent A but it is insoluble in solvent B. If solvents A and B are miscible, will the organic compound be soluble in a mixture of A and B?. How might this be used to purify the organic compound?

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Anu M.

The organic compound soluble in solvent A but insoluble in solvent B may or may not be soluble in a mixture of A and B depending upon the solvent ratio. The separation is based on “Like dissolves like”. We will purify the organic compound by following method. The component mixture is dissolved in a suitable solvent A and a second solvent B that is miscible with the first solvent is added until the organic compound separates out. Next, the contents are thoroughly mixed (shaking) and allowed to settle. The solvent mixture and the compound separate out. During the process the impurity gets dissolved in the solvent mixture. The solvent mixture containing impurity is separated by decanting to give the pure organic compound

### Subject:Chemistry

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Question:

Explain in terms of a dynamic equilibrium why the smaller crystals are observed to dissolve while the larger ones grow even larger in a saturated solution solution kept for a period of time .

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Anu M.

A process is said to be in dynamic equilibrium when rate in the forward direction is equal to the rate of the reverse direction. The conditions that are necessary to achieve equilibrium in physical and chemical systems are • Constant observable macroscopic properties (e.g., temperature, pressure, concentration), • A closed system, • Reversibility of the change. • Equal rates of opposing change. In a saturated solution there is a dynamic equilibrium between dissolved solute and the undissolved solute particles. Solute enters the solution at the same rate as the solute in the solution precipitates back. Small crystal has large surface to volume ratio as compared to that of large crystal. Because of this factor, small crystals dissolve more in amount than large crystals. Thus the dynamic equilibrium makes smallest of the crystals to dissolve more compared to large crystals On the other hand, dissolved solute deposits back on the undissolved solute at the same rate in both cases. Thus, consequently and eventually for many small crystals in contact with a saturated solution, the smallest of the crystals are observed to dissolve while the larger ones grow even larger.

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