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JEE Advanced Mix Test 24
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JEE Advanced Mix Test 24
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  • Question 1/10
    4 / -1

    The enthalpy in the process HCl + nH2O → HCl in n moles of H2O where n is the number of moles of water, is called the integral heat of solution. When n is large enough that continued addition of water does not increase the heat of solution, one simply writes HCl + aq → HCl (aq) The enthalpy for this process is the limiting value for the integral heat of solution. The enthalpy for the process HCl in n moles of H2O + m H2O → HCl in (n + m) moles of H2O is called the integral heat of dilution. These quantities are indicated in figure. Another quantity of interest is the differential heat of solution, defined as the slope of the enthalpy curve. The heats of solution depend on the composition of the solution as shown in the figure

    Integral heat of solution for the following step is :

    HCl + 5H2O → HCl (5H2O)

    Solutions

    From the graph.

  • Question 2/10
    4 / -1

    The enthalpy in the process HCl + nH2O → HCl in n moles of H2O where n is the number of moles of water, is called the integral heat of solution. When n is large enough that continued addition of water does not increase the heat of solution, one simply writes HCl + aq → HCl (aq) The enthalpy for this process is the limiting value for the integral heat of solution. The enthalpy for the process HCl in n moles of H2O + m H2O → HCl in (n + m) moles of H2O is called the integral heat of dilution. These quantities are indicated in figure. Another quantity of interest is the differential heat of solution, defined as the slope of the enthalpy curve. The heats of solution depend on the composition of the solution as shown in the figure.

    What is the approximate enthalpy change for the reaction HCl + aq →HCl (aq)

    Solutions

    From the graph.

  • Question 3/10
    4 / -1

    The enthalpy in the process HCl + nH2O → HCl in n moles of H2O where n is the number of moles of water, is called the integral heat of solution. When n is large enough that continued addition of water does not increase the heat of solution, one simply writes HCl + aq → HCl (aq) The enthalpy for this process is the limiting value for the integral heat of solution. The enthalpy for the process HCl in n moles of H2O + m H2O → HCl in (n + m) moles of H2O is called the integral heat of dilution. These quantities are indicated in figure. Another quantity of interest is the differential heat of solution, defined as the slope of the enthalpy curve. The heats of solution depend on the composition of the solution as shown in the figure.

    Approximate value of differential heat of that solution in which 1 mole of HCl is dissolved in 6 mole of water is:

    Solutions

    From the graph the slope of line shown in graph

  • Question 4/10
    4 / -1

    The CO in a 20 L sample of gas was converted to CO2 by passing the gas over iodine pentoxide heated to 150ºC:
                        I2O5(s) + 5CO(g) → 5CO2(g) + I2(g) .
        The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M Na2S2O3. The excess hypo was back-titrated with 5 mL of 0.001 M I2 solution.

    What must be the milligrams of CO in 1 L of the original gas sample?

    Solutions

  • Question 5/10
    4 / -1

    The CO in a 20 L sample of gas was converted to CO2 by passing the gas over iodine pentoxide heated to 150ºC:
    I2O5(s) + 5CO(g) → 5CO2(g) + I2(g) .
    The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M Na2S2O3. The excess hypo was back-titrated with 5 mL of 0.001 M I2 solution.

    What is the weight ( in mg) of iodine produced due to the reaction with I2O5(s) when it reacted with CO(g) in 20 L sample of gas?

    Solutions

  • Question 6/10
    4 / -1

    The CO in a 20 L sample of gas was converted to CO2 by passing the gas over iodine pentoxide heated to 150ºC:

    I2O5(s) + 5CO(g) → 5CO2(g) + I2(g).

    The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M Na2S2O3. The excess hypo was back-titrated with 5 mL of 0.001 M I2 solution.

    In the above reaction (given in the comprehension) CO(g) is acting as

    Solutions

  • Question 7/10
    4 / -1

    Solutions

  • Question 8/10
    4 / -1

    Solutions

  • Question 9/10
    4 / -1

    Solutions

  • Question 10/10
    4 / -1

    Solutions

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