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Chemistry Test - 48

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Chemistry Test - 48
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  • Question 1
    1 / -0

    The most acidic oxyacid of halogen among the given option is

    Solution

    The acidity of oxyacids of halogen can be compared by checking the oxidation state of halogen in the acid. Higher the oxidation state, higher the acidity. Here H5IO6 has an oxidation state of +7 and hence is the strongest acid.

  • Question 2
    1 / -0

    50cm3 of 0.04MK2Cr2O7 in acidic medium oxidizes a sample of H2S gas to sulphur. Volume of 0.03MKMnO4required to oxidize the same amount of H2S gas to sulphur, in acidic medium is :

    Solution
    Equivalents of \(K_{2} C r_{2} O_{7}=\) Equivalents of \(H_{2} S=\) Equivalents of \(K M n O_{4}\)
    \(n_{f}\) of \(K_{2} C r_{2} O_{7}=6\)
    \(n_{f}\) of \(K M n O_{4}=5\)
    Equivalents \(=n_{f} \times\) Molarity \(\times\) Volume
    \(6 \times 0.04 \times 50=5 \times 0.03 \times V_{K M n O_{4}}\)
    \(V_{K M n O_{4}}=80 \mathrm{cm}^{3}\)
  • Question 3
    1 / -0

    Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the following statements is incorrect for these reactions?

    Solution
    The reactions that take place when excess of \(K I\) reacts with \(C u S O_{4}\) and then adding \(N a_{2} S_{2} O_{3}\) are shown below:
    \(4 K I+2 C u S O_{4} \rightarrow I_{2}+C u_{2} I_{2}+2 K_{2} S O_{4}\)
    \(I_{2}+2 N a_{2} S_{2} O_{3} \rightarrow N a_{2} S_{4} O_{6}+2 N a I\)
    During this process, \(C u_{2} I_{2}\) is formed while \(C u I_{2}\) is not formed. \(N a_{2} S_{2} O_{3}\) reduces the liberated iodine. In the second reaction, the oxidation state of iodine changes from 0 to \(-1,\) while that of sulphur changes from +2 to +2.5
  • Question 4
    1 / -0

    A two litre flask contains 22 g of carbon dioxide and 1 g of helium at 200 C. Calculate the partial pressure exerted by CO2 and He if the total pressure is 3 atm.

    Solution
    The molar masses of carbon dioxide and helium are \(44 \mathrm{g} / \mathrm{mol}\) and \(4 \mathrm{g} / \mathrm{mol}\) respectively. The number of moles is the ratio of mass to molar mass.
    \(22 \mathrm{g}\) of carbon dioxide \(=\frac{22}{44}=0.5 \mathrm{moles}\)
    \(1 \mathrm{g}\) of helium \(=\frac{1}{4}=0.25 \mathrm{moles}\)
    The mole fraction of carbon dioxide \(=\frac{0.5}{0.5+0.25}=0.667\)
    The mole fraction of helium \(=\frac{0.25}{0.5+0.25}=0.333\)
    The partial pressure of a gas is the product of its mole fraction and total pressure. The partial pressure of carbon dioxide \(=0.667 \times 3=2 \mathrm{atm}\)
    The partial pressure of helium \(=0.333 \times 3=1\) atm.
  • Question 5
    1 / -0

    3.6 gm of an ideal gas was injected into a bulb of internal volume of 8 L at pressure P atm and temp T − K. The bulb was then placed in a thermostat maintained at (T+15) K.0.6 gm of the gas was

    let off to keep the original pressure. Find P and T if mol weight of gas is 44.

    Solution
    \(n=\frac{3.6}{44}=0.0818\)
    \(n^{\prime}=\frac{3.6-0.6}{44}=0.06818\)
    since \(P\) and \(V\) are constant, \(n T=n^{\prime} T^{\prime}\) \(0.0818 T=0.06818 T+1.0227\)
    \(T=75 K\)
    \(P=\frac{n R T}{V}=\frac{0.0818 \times 75 \times 0.0821}{8}=0.062 \mathrm{atm}\)
  • Question 6
    1 / -0

    Children brought some balloons each of 2 litre capacity to a chemist and asked him to fill them with hydrogen gas. The chemist possessed 8 litre cylinder containing hydrogen at 10 atm pressure at room temperature. How many balloons could he fill with hydrogen gas at normal atmospheric pressure at the same temperature?

    Solution
    Initial pressure in the cylinder \(P=10\) atm. Initial volume in the cylinder \(V=8 \mathrm{L}\). Final pressure in the balloons \(P^{\prime}=1\) atm. Final volume in the balloons \(V^{\prime}=? ?\) \(P V=P^{\prime} V^{\prime}\) at same temperature 10 atm \(\times 8 \mathrm{L}=1\) atm \(\times V^{\prime}\)
    \(V^{\prime}=80 \mathrm{L}\)
    The number of balloons ( each of 2 litre capacity ) that he could fill
    \(=\frac{80 \mathrm{L}}{2 \mathrm{L}}=40\)
  • Question 7
    1 / -0

    While resting, the average human male use 0.2 dm3 of O2 per hour at 1 atm & 273 K for each kg of body mass. Assume that all this O2 is used to produce energy by oxidising glucose in the body. What is the mass of glucose required per hour by a resting male having mass 60 kg. What volume, at 1 atm and 273 K of CO2 would be produced?

    Solution
    \(C_{6} H_{12} O_{6}+6 O_{2} \rightarrow 6 C O_{2}+6 H_{2} O\)
    \(n_{O_{2}}=60 \times \frac{0.2}{22.4}=0.5357\)
    \(n_{g l u c o s e}=\frac{0.5357}{6}=0.08928\)
    \(W_{g l u c o s e}=180 \times 0.08928=16.07\)
    \(n_{C O_{2}}=n_{O_{2}}=0.5357\)
    \(V_{C O_{2}}=22.4 \times 0.5357=12 d m^{3}\)
  • Question 8
    1 / -0

    The mass of potassium dichromate crystals required to oixidise 750 cm3 of 0.6 M Mohr's salt solution is : (Given, molar mass, potassium dichromate = 294, Mohr's salt = 392)

    Solution
    Amount of Mohr's salt oxidized is \(0.75 \mathrm{mL} \times 0.6 \mathrm{mol} / l=0.45 \mathrm{mol}\)
    \(K_{2} C r_{2} O_{7}+6 F e S O_{4}+7 H_{2} S O_{4} \rightarrow K_{2} S O_{4}+C r_{2}\left(S O_{4}\right)_{3}+3 F e_{2}\left(S O_{4}\right)_{3}+7 H_{2} O\)
    For 6 mol of Mohr's salt 1 mol of Pottasium dichromate is required For 0.45 mol of Mohr's salt, Pottasium dichromate required is \(\frac{1 \times 0.45}{6}=0.075 \mathrm{mol}\) Mass of potassium dichromate required is \(0.075 \times 294=22.05 g\)
  • Question 9
    1 / -0

    At a certain temperature the time required for the complete diffusion of 200 ml of H2 gas is 30 min. The time required for the complete diffusion of 50 ml of O2 gas at the same temperature will be:

    Solution
    According to Graham's law of diffusion,
    Rate of diffusion, \(\Gamma \propto \frac{1}{\sqrt{M}}=\frac{V}{t}\)
    [Here, \(M=\) molecular mass, \(V=\) volume and \(t=\) time \(]\) Thus, for \(H_{2}\) gas,
    \(\frac{200}{30}=\frac{1}{\sqrt{2}}\)
    For \(O_{2}\) gas,
    \(\frac{50}{t}=\frac{1}{\sqrt{32}}\)
    \(\ldots \ldots . .(\) ii \()\)
    From equations (i) and (ii), we get
    \(\frac{200 \times t}{30 \times 50}=\frac{\sqrt{32}}{\sqrt{2}}\)
    \(t=\frac{\sqrt{16} \times 30 \times 50}{200}\)
    \(=\frac{4 \times 30}{4}=30 \mathrm{min}\)
  • Question 10
    1 / -0

    If 20 mL of 0.1 M K2Cr2O7 is required to titrate 10 mL of a liquid iron supplement, then the concentration of iron in the the the vitamin solution is :

    Solution
    The redox reaction is as follows:
    \(C r_{2} O_{7}^{2-}+F e^{2+}+14 H^{+} \rightarrow 2 C r^{3+}+F e^{3+}+7 H_{2} O\)
    The redox changes involved are given below:
    \(\mathrm{i} .6 e^{-}+C r_{2} O_{7}^{2-} \rightarrow 2 C r^{3+}(x=6)\)
    ii. \(F e^{2+} \rightarrow F e^{3+}+e^{-}(x=1)\)
    Milliequivalents of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}=\) Milliequivalents of \(\mathrm{Fe}^{2+}\) \(20 \times 0.1 \times 6=10 \times N\)
    \(N=1.2\)
    \(M=\frac{N}{{ }^{\prime} n^{\prime} f a c t o r}=\frac{1.2}{1}=1.2 M\)
    Hence, the concentration of iron in the vitamin solution is \(1.2 \mathrm{M}\).
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