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CBSE 12th Standard Chemistry Subject Solution Case Study Questions With Solution 2021
By QB365 on 21 May, 2021
QB365 Provides the updated CASE Study Questions for Class 12 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams
QB365 - Question Bank Software
12th Standard CBSE
Final Semester - June 2015
Read the passage given below and answer the following questions: The concentration of a solute is very important in studying chemical reactions because it determines how often molecules collide in solution and thus indirectly determine the rate of reactions and the conditions at equilibrium. There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. Concentration can be expressed in terms of molarity, molality, parts per million, mass percentage, volume percentage, etc. The following questions are multiple choice questions. Choose the most appropriate answer: (i) The molarity (in mol L -1 ) of the given solution will be
(ii) Which of the following is correct relationship between mole fraction and molality?
(iii) Which of the following is temperature dependent?
(iv) Which of the following is true for an aqueous solution of the solute in terms of concentration?
Read the passage given below and answer the following questions: At 298 K, the vapour pressure of pure benzene, C 6 H 6 is 0.256 bar and the vapour pressure of pure toluene C 6 H 5 CH 3 is 0.0925 bar. Two mixtures were prepared as follows: (i) 7.8 g of C 6 H 6 + 9.2 g of toluene (ii) 3.9 g of C 6 H 6 + 13.8 g of toluene The following questions are multiple choice questions. Choose the most appropriate answer: (i) The total vapour pressure (bar) of solution 1 is
(ii) Which of the given solutions have higher vapour pressure?
(iii) Mole fraction of benzene in vapour phase in solution 1 is
(iv) Solution I is an example of a/an
Read the passage given below and answer the following questions: An ideal solution may be defined as the solution which obeys Raoult's law exactly over the entire range of concentration. The solutions for which vapour pressure is either higher or lower than that predicted by Raoult's law are called non-ideal solutions. Non-ideal solutions can show either positive or negative deviations from Raoult's law depending on whether the A-B interactions in solution are stronger or weaker than A - A and B - B interactions. The following questions are multiple choice questions. Choose the most appropriate answer: (i) Which of the following solutions is/are ideal solution(s)? (i) Bromoethane and iodoethane (ii) Acetone and chloroform (iii) Benzene and acetone (iv)n-heptane and n-hexane
(ii) Which of the following is not true for positive deviations?
(iii) For water and nitric acid mixture which of the given graph is correct?
(iv) Water- HCl mixture I. shows positive deviations II. forms minimum boiling azeotrope III. shows negative deviations IV. forms maximum boiling azeotrope
Read the passage given below and answer the following questions: The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties. Relative lowering in vapour pressure is also an example of colligative properties. For an experiment, sugar solution is prepared for which lowering in vapour pressure was found to be 0.061 mm of Hg. (Vapour pressure of water at 20°C is 17.5 mm of Hg.) The following questions are multiple choice questions. Choose the most appropriate answer: (i) Relative lowering of vapour pressure for the given solution is
(ii) The vapour pressure (mm of Hg) of solution will be
(iii) Mole fraction of sugar in the solution is
(iv) The vapour pressure (mm of Hg) of water at 293 K when 25 g of glucose is dissolved in 450 g of water is
Read the passage given below and answer the following questions: Few colligative properties are: (a) relative lowering of vapour pressure: depends only on molar concentration of solute (mole fraction) and independent of its nature. (b) depression in freezing point: it is proportional to the molal concentration of solution. (c) elevation of boiling point: it is proportional to the molal concentration of solute. (d) osmotic pressure: it is proportional to the molar concentration of solute. A solution of glucose is prepared with 0.052 g at glucose in 80.2 g of water. (K f = 1.86 K kg mol -1 and K b = 5.2 K kg mol -1 ) The following questions are multiple choice questions. Choose the most appropriate answer: (i) Molality of the given solution is
(ii) Boiling point for the solution will be
(iii) The depression in freezing point of solution will be
(iv) Mole fraction of glucose in the given solution is
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Cbse 12th standard chemistry subject solution case study questions with solution 2021 answer keys.
(i) (d) : Density of solution = 1.202 g/mL Volume of solution = \(\frac{100 \mathrm{~g}}{1.202 \mathrm{~g} / \mathrm{mL}}=83.2 \mathrm{~mL}\) Molarity = \(\frac{n_{\mathrm{KI}}}{\text { Volume of solution in } \mathrm{L}}\) \(=\frac{0.120 \mathrm{~mol}}{0.0832 \mathrm{~L}}=1.4423 \mathrm{~mol} \mathrm{~L}^{-1}\) (ii) (a): \(x_{2}=\frac{n_{2}}{n_{1}+n_{2}} ; x_{1}=\frac{n_{1}}{n_{1}+n_{2}} ; \frac{x_{2}}{x_{1}}=\frac{n_{2}}{n_{1}}\) \(\frac{x_{2}}{x_{1}}=\frac{m_{2} / M_{2}}{m_{1} / M_{1}}=\frac{m_{2}}{m_{1}} \times \frac{M_{1}}{M_{2}}\) ...(i) Molality = \(\frac{n_{2}}{m_{1}}=\frac{m_{2}}{M_{2} \times m_{1}}\) ...(ii) From(i) and (ii), m = \(\frac{x_{2}}{x_{1}} \times \frac{1}{M_{1}} ; x_{1}=1-x_{2}\) Hence. x 2 = \(\frac{m M_{1}}{1+m M_{1}}\) (iii) (a) : Mass does not depend on temperature while volume does. Hence, molarity depends on temperature. (iv) (b): 1M solution contains 1 mole of solute in less than 1000 g of the solvent whereas 1 m solution has 1 mole of the solute in 1000 g of the solvent.
(i) (b) : Moles of C 6 H 6 = \(\frac{7.8}{78}=0.1\) Mole C 6 H 5 CH 3 = \(\frac{9.2}{92}=0.1\) Mole fraction of C 6 H 6 = \(\frac{0.1}{0.1+0.1}=0.5\) => Mole fraction of C 6 H 5 CH 3 = 0.5 Vapour pressure of toluene = Vapour pressure of pure toluene x mole fraction of toluene = 0.0925 x 0.5 = 0.04625 Vapour pressure of benzene = 0.256 x 0.5 = 0.128 Total vapour pressure of solution = 0.17425 (ii) (a) : Moles of benzene in solution-II = \(\frac{3.9}{78}=0.05\) Moles of toluene in solution-II = \(\frac{13.8}{92}=0.15\) Vapour pressure of solution = 0.256 x 0.05 + 0.0925 x 0.15 = 0.0128 + 0.013875 = 0.026675 (iii) (c) : Mole fraction of benzene in vapour phase \(y_{\text {benzene }}=\frac{p_{\text {benzene }}}{P_{\text {total }}}=\frac{0.128}{0.17425}=0.734\) (iv) (a) : Benzene and toluene form an ideal solution.
(i) (d) : II represents negative deviations and III represents positive deviations. (ii) (b): For positive deviations \(p_{A}>p_{A}^{\circ} x_{A} \text { and } p_{B}>p_{B}^{\circ} x_{B}\) (iii) (b) : Water and nitric acid mixture shows negative deviations from Raoult's law, hence \(p_{A}<p_{A}^{\circ} x_{A} \text { and } p_{B}<p_{B}^{\circ} x_{B}\) (iv) (d): Water-HCl mixture shows negative deviations from Raoult's law and solutions showing negative deviations from ideal behaviour form maximum boiling azeotrope.
(i) (a) : Vapour pressure of water \(\left(p_{A}^{\circ}\right)\) = 17.5 mm of Hg Lowering of vapour pressure \(\left(p_{A}^{\circ}-p_{A}\right)\) = 0.061 Relative lowering of vapour pressure \(=\frac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}}=\frac{0.061}{17.5}=0.00348\) (ii) (c): P = Vapour pressure of solvent - lowering in vapour pressure = 17.5 - 0.061 = 17.439 mm of Hg (iii) (a): \(\frac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}}=x_{B}=0.00348\) Hence, mole fraction of sugar = 0.00348 (iv) (b): \(\frac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}}=x_{B}=\frac{w_{B} \times M_{A}}{M_{B} \times w_{A}}\) \(\frac{17.5-p_{A}}{17.5}=\frac{25 \times 18}{450 \times 180}=5.56 \times 10^{-3}\) \(17.5-p_{A}=17.5 \times 5.56 \times 10^{-3}\) \(17.5-p_{A}=0.0973\) P = 17.40 mm Hg
(i) (b) : m \(=\frac{0.052}{180} \times \frac{1000}{80.2}=0.0036\) (ii) (c): \(\Delta T_{b}=K_{b} \times m=5.2 \times 0.0036=0.0187 \mathrm{~K}\) \(T_{b}=373+0.0187=373.0187 \mathrm{~K} \approx 373.02 \mathrm{~K}\) (iii) (d): \(\Delta T_{f}=K_{f} \times m=1.86 \times 0.0036=0.067 \mathrm{~K}\) (iv) (a): Moles of glucose \(=\frac{0.052}{180}=0.00028\) Moles 0f water = \(\frac{80.2}{18}=4.455\) Mole fraction of glucose = \(\frac{0.00028}{4.45+0.00028}=6.28 \times 10^{-5}\)
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CBSE Board Class 12 Chemistry Case Based Questions With Solutions
CBSE Board Class 12 Chemistry Case Based Questions With Solutions to prepare for the upcoming term 1 board exams are given here. It is extremely helpful since the expert team has crafted this by following the Syllabus of class 12th. Case study questions are also known as passage based problems because it includes the paragraph from which 5 or more problems are curated.
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