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Chemistry Test - 6

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Chemistry Test - 6
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  • Question 1
    1 / -0

    A,B and C gases taken in a volume ratio of 3:2:4 contains atoms in the ratio of 1:2:8/3. The gases having minimum and maximum atomicity are respectively:

    Solution
    Ratio of volumes for gases \(A: B: C=3: 2: 4\)
    since, volume of a gas is directly proportional to moles at a given set of conditions,
    ratio of moles of gases \(=3: 2: 4\)
    Ratio of atoms \(=1: 2: 8 / 3\)
    3 moles of gas \(A\) contain atoms \(=x\)
    1 molecule of gas has atoms \(=\frac{x}{3 N_{A}}\)
    Atomicity \(=\frac{x}{3 N_{A}}\)
    2 moles of gas \(B\) contain atoms \(=2 x\)
    1 molecule of gas has atoms \(=\frac{2 x}{2 N_{A}}\)
    Atomicity \(=\frac{x}{N_{A}}\)
    4 moles of gas \(C\) contain atoms \(=\frac{8}{3} x\)
    1 molecule of gas has atoms \(=\frac{\frac{8 x}{3}}{4 N_{A}}\)
    Atomicity \(=\frac{2 x}{3 N_{A}}\)
    Therefore gas \(A\) has minimum atomicity while gas \(B\) has maximum atomicity.
    Hence option C is the correct answer.
  • Question 2
    1 / -0

    Which of the following are arranged from smallest unit to largest unit?

    Solution

    mm<cm<m<km

    1 kilometer =1000000 millimeter

    1 meter =1000 millimeter

    1 centimeter =10 millimeter

    Hence, option D is correct.

  • Question 3
    1 / -0

    A triatomic gas having 3.6132×1024 atoms has to be stored in a container.What should be the volume of container to store the gas at 1 atm pressure?

    Solution
    Number of atoms in Triatomic gas \(=3.6132 \times 10^{24}\) atoms
    Number of triatomic molecules in container \(=\frac{3.6132 \times 10^{24}}{3}=1.2044 \times 10^{24}\)
    Number of moles \(=\frac{1.2044 \times 10^{24}}{6.023 \times 10^{23}}=2 moles\)
    Volume of 1 mole gas \(=22.4 L\) at
    STP Volume of 2 mole gas \(=22.4 \times 2=44.8 L\)
    Hence option C is the correct answer.
  • Question 4
    1 / -0

    The number of significant figures in 0.00052 is :

    Solution

    Zeros preceding the first non-zero digit are not significant. This zero indicates the position of the decimal point. Thus, 0.00052 contains two significant figures.
    Hence option B is correct.

  • Question 5
    1 / -0

    The density of mercury is 13.6 g/mL. What is the volume of a 200 gram sample of mercury? (Hint: use the density as a conversion factor.)

    Solution
    The density of mercury is \(13.6 g / lm\). The volume of a \(200 gram\) sample of mercury will be :
    Volume\(=\frac{200 g }{13.6 g / mL }\)
    Volume\(=14.7 mL\)
    Hence option D is the correct answer.
  • Question 6
    1 / -0

    Significant figures in a number show:

    Solution

    The significant figures present in a number are digits with reasonable certainty. The number may also contain non significant figures which are digits with uncertainty.
    Hence option A is correct.

  • Question 7
    1 / -0

    112 litres of a gas contains 1.2044×1025atoms. The gas can be:

    Solution
    \(112 L\) of gas \(=\frac{112}{22.4}=5 molegas\)
    5 mole gas molecules has \(=1.2044 \times 10^{25}\) atoms
    Atoms in 1 mole \(=\frac{1.2044 \times 10^{25}}{5}\)
    Atoms in 1 molecule of gas \(=\frac{1.2044 \times 10^{25}}{5 \times 6.023 \times 10^{23}}=4\)
    In the given options tetra-atomic gas is \(N H_{3.}\)
    Hence option B is the correct answer.
  • Question 8
    1 / -0

    The number of nucleons present in 2.4 mL of CO2 at STP are:

    Solution
    Nucleons are the proton and the neutron, which are the constituents of the atomic nucleus nucleons are the proton and the neutron, which are the constituents of the atomic nucleus. For each molecule of \(C O_{2}\) there is 1 atom of \(C\) which has 6 proton and 6 neutrons.
    2 atoms of \(O\) which has 8 proton and 8 neutrons.
    Total number of nucleons in 1 molecule of \(C O_{2}=12+32=44\)
    Moles of \(C O_{2}\) in \(2.4 m l=\frac{2.4 \times 10^{-3}}{22.4}=1.07 \times 10^{-4}\) moles
    Number of nucleons \(=44 \times 1.07 \times 10^{-4} \times 6.023 \times 10^{23}=2.648 \times 10^{21}\)
    Hence option C is the correct answer.
  • Question 9
    1 / -0

    While having a fire drill session in Toppr, 11.2L of CO2 gas was used at STP. Now, Ritika is curious to know the total number of oxygen atoms spent in the fire drill. She provides the above data to an intern to get her the answer. The answer is:

    Solution
    Number of moles in \(11.2 L\) of \(C O_{2}=\frac{11.2}{22.4}=0.5\) moles
    In one molecule of \(C O_{2}\) number of oxygen atom \(=2\) atoms
    \(\Rightarrow\) Number of oxygen moles in 0.5 mole \(C O_{2}=2 \times 0.5=1\) mole
    Number of atoms in 1 mole \(=\) Avogadro's number \(=6.023 \times 10^{23}\) atoms
    Hence option D is the correct answer.
  • Question 10
    1 / -0

    Calculate the number of moles of electrons present in 11.2L of N2.

    Solution
    Number of moles of \(N_{2}\) in \(11.2 L=\frac{11.2}{22.4}=0.5\) moles
    Number of atoms in 1 molecule of \(N_{2}=2\)
    Number of atoms in 0.5 mole \(N_{2}=2 \times 0.5=1\) mole
    Number of electrons per atom of \(N=7\)
    Number of electrons in 1 mole atoms \(=7 \times N_{A}\)
    Hence option B is the correct answer.
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