6- What is the freezing point of an aqueous solution that boils at 105.docx

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6. What is the freezing point of an aqueous solution that boils at 105.0 oC? Solution The boiling point of an aqueous solution is 105.0Â°C. The boiling point of pure water is 100.0Â°C. The elevation of boiling point is given as ?T b = (boiling point of solution) â€“ (boiling point of pure water) = (105.0Â°C) â€“ (100.0Â°C) = 5.0Â°C. As per Raoultâ€™s law, we have, ?T b = K b *(molality of the solution) where K b = 0.512Â°C/m is the boiling point elevation constant of water. Plug in values and obtain 5.0Â°C = (0.512Â°C/m)*(molality of the solution) ====> molality of the solution = (5.0Â°C)/(0.512Â°C/m) = 9.7656 m. Assume that the molality of the solution remains constant; therefore, we can find the freezing point depression of the aqueous solution using Raoultâ€™s law as below. ?T f = K f *(molality of the solution) where K f = freezing point depression constant of water = 1.86Â°C/m. Plug in values and get ?T f = (1.86Â°C/m)*(9.7656 m) = 18.1640Â°C. We define ?T f = (T f ) pure â€“ (T f ) solution where T f,pure = 0.0Â°C is the freezing point of pure water and T f,solution is the freezing point of the aqueous solution. Plug in values and get 18.1640Â°C = 0.0Â°C â€“ T f , solution ====> T f,solution = 0.0Â°C â€“ 18.1640Â°C = -18.1640Â°C ? -18.2Â°C (ans). .

6. What is the freezing point of an aqueous solution that boils at 105.0 oC?
Solution
The boiling point of an aqueous solution is 105.0Â°C. The boiling point of pure water is
100.0Â°C. The elevation of boiling point is given as
?T b = (boiling point of solution) â€“ (boiling point of pure water) = (105.0Â°C) â€“ (100.0Â°C)
= 5.0Â°C.
As per Raoultâ€™s law, we have,
?T b = K b *(molality of the solution)
where K b = 0.512Â°C/m is the boiling point elevation constant of water. Plug in values and
obtain
5.0Â°C = (0.512Â°C/m)*(molality of the solution)
====> molality of the solution = (5.0Â°C)/(0.512Â°C/m) = 9.7656 m.
Assume that the molality of the solution remains constant; therefore, we can find the freezing
point depression of the aqueous solution using Raoultâ€™s law as below.
?T f = K f *(molality of the solution)
where K f = freezing point depression constant of water = 1.86Â°C/m. Plug in values and get
?T f = (1.86Â°C/m)*(9.7656 m) = 18.1640Â°C.
We define ?T f = (T f ) pure â€“ (T f ) solution
where T f,pure = 0.0Â°C is the freezing point of pure water and T f,solution is the freezing point of the
aqueous solution. Plug in values and get
18.1640Â°C = 0.0Â°C â€“ T f , solution

====> T f,solution = 0.0Â°C â€“ 18.1640Â°C = -18.1640Â°C ? -18.2Â°C (ans).

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Both these questions go together Solution When we bring a magnet near a metal, the metal enters a magnetic field, all of the electrons inside the metal \"line up,\" causing a temporary magnetic alignment that is attracted to the magnet . That alignment dissipates once the magnetic field is removed and therefore, the only way for a metal to be repelled by a magnet is if it\'s first magnetized to the opposite pole. so if the metal is magnetised and the a magnet with the same pole is brought near it will repel Note:- its easy to magnetise a iron piece just rub the magnet over the metal in one direction many, many times. By rubbing a magnet over your metal continuously, you are aligning the electrons in the metal in a certain way, toward a certain polarity. The more that you do this, the longer the effect will last, taking more time for the electrons to return to their random or unpolarized state. so the answer to the question is 1) c 2) d .

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6. Both solutions and colloids are mixtures. Differentiate between them. Solution Although solutions and colloid both are mixtures but we can differentiate solution and colloid on the basis of some parameter given below: 1) Particle size: Particle size of solution is less than 1nm or (10 -9 ) m Particles size of colloid are between 1nm to 1000 nm 2) Visibility In solutions the visibility of substances dissolved in it are not possible even with optical means. While in colloid it may be visible with ultra microscope. 3) Diffusion Diffusion is very rapid in solution while very slow in colloid 4) setting Solution doesn\'t settle while colloid may settle down under centrifuge . 5) Nature: Solution is Homogeneous in nature while colloid is heterogeneous in nature. 6) Tyndall effect Solution doesn\'t show Tyndall effect Colloid shows Tyndall effect So we can say that solution and colloid both are mixture but there are a lot of difference between them. .

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6). The Town of Elmwood applied for a competitive grant from the state government for providing an after- school recreational program for at-risk children from low-income families. On 2/1/2005, Elmwood was notified that it had been awarded a grant of $100,000 for the program. The program provides up to $1,000 per child served provided that family annual income is less than $25,000. The grant provides for reimbursement of all allowable costs of conducting the program on a twice-monthly basis. No time limit is imposed for completion of the program. Any unused grant award may be carried forward to the next fiscal year. By 12/31/2005, Elmwood?s fiscal year-end, $74,000 had been expended in providing recreational services for eligible beneficiaries. Reimbursement of $71,000 had been received from the state by this date. Required: [12 pts.] a). What are the eligibility requirements of this grant? b). As of 12/31/2005, identify the following: (1). Cash Received (2). Program Revenue (3). Expenditures Solution Total Expenditure = $ 74000 Cash Received = $ 71000 Revenue = 71000-74000 = $ 3000 The program must be carried on & also the expenditure will be for the children whose family income is less than $ 25000/-. .

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500 men are arranged in an array of 10 rows and 50 columns according to their heights. Tallest among each row of all are asked to come out. And the shortest among them is A. Similarly after resuming them to their original positions, the shortest among each column are asked to come out. And the tallest among them is B. Now who is taller A or B ? Solution No one is taller, both are same as A and B are the same person. .

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1. 6. What is the freezing point of an aqueous solution that boils at 105.0 oC? Solution The boiling point of an aqueous solution is 105.0Â°C. The boiling point of pure water is 100.0Â°C. The elevation of boiling point is given as ?T b = (boiling point of solution) â€“ (boiling point of pure water) = (105.0Â°C) â€“ (100.0Â°C) = 5.0Â°C. As per Raoultâ€™s law, we have, ?T b = K b *(molality of the solution) where K b = 0.512Â°C/m is the boiling point elevation constant of water. Plug in values and obtain 5.0Â°C = (0.512Â°C/m)*(molality of the solution) ====> molality of the solution = (5.0Â°C)/(0.512Â°C/m) = 9.7656 m. Assume that the molality of the solution remains constant; therefore, we can find the freezing point depression of the aqueous solution using Raoultâ€™s law as below. ?T f = K f *(molality of the solution) where K f = freezing point depression constant of water = 1.86Â°C/m. Plug in values and get ?T f = (1.86Â°C/m)*(9.7656 m) = 18.1640Â°C. We define ?T f = (T f ) pure â€“ (T f ) solution where T f,pure = 0.0Â°C is the freezing point of pure water and T f,solution is the freezing point of the aqueous solution. Plug in values and get 18.1640Â°C = 0.0Â°C â€“ T f , solution

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