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Some Basic Concepts of Chemistry Test - 68

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Some Basic Concepts of Chemistry Test - 68
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  • Question 1
    1 / -0
    Two beakers AA and BB present in a closed vessel. Beaker AA contains 152.4g152.4\:g aqueous solution of urea, containing  12g12\:g of urea. Beaker BB contains 196.2g196.2\:g glucose solution, containing 18g18\:g of glucose. Both the solutions allowed to attain the equilibrium. Determine weight %\% of glucose in its solution is at equilibrium.
    Solution
    The molar masses of glucose and urea are 180 g/mol and 60 g/mol respectively.
    The number of moles of urea are 1260=0.2\dfrac {12}{60} = 0.2
    The number of moles of glucose are 18180=0.1\dfrac {18}{180} = 0.1
    When equilibrium is achieved, the mass (196.2 g196.2 \ g) of glucose solution will reduce by x g and the mass of urea (152.4 g152.4 \ g) will increase by x g. Due to this, the concentration (ratio of number of moles of solute to the mass of solution) of urea solution will become equal to the concentration of glucose solution.
    $$\displaystyle  \frac {0.2}{152.4+x} = \frac {0.1}{196.2-x} \\
    152.4+x= 392.4-2x \\
    3x = 240 \\
    x = 80 $$
    Total mass of glucose solution  =196.280=116.2\displaystyle  = 196.2-80 = 116.2 g
    weight percent of glucose =18116.2×100=15.5\displaystyle = \frac {18}{116.2} \times 100 =15.5 %
  • Question 2
    1 / -0
    The molality of 11 L solution with x%x\% H2SO4H_2SO_4 is 99. The weight of the solvent present in the solution is 910910 g. The value of xx is:
    Solution
    11 L of solution contains 99 moles of sulfuric acid which corresponds to 9×98=8829 \times 98 = 882 g.
    11 L of solution contains 910910 g water.
    Hence, the percentage of sulphuric acid in the solution is 882882+910×100=49.2\dfrac {882}{882+910}\times 100 = 49.2%.
  • Question 3
    1 / -0
    n1\displaystyle n_{1} g of substance X reacts with n2\displaystyle n_{2} g of substance Y to form m1\displaystyle m_{1} g of substance R and m2\displaystyle m_{2} g of substance of S. This reaction can be represented as X+YR+SX + Y \rightarrow R + S. The relation which can be established in the amounts of the reactants and the products will be :
    Solution
    The relation which can be established in the amounts of the reactants and the products will be as follows:

    Total mass of reactants=Total mass of products\displaystyle \text {Total mass of reactants}=\text {Total mass of products}

    n1+n2=m1+m2\displaystyle n_{1}+n_{2}=m_{1}+m_{2}

    This is based on the law of conservation of mass.
  • Question 4
    1 / -0
    Na2SO4.xH2ONa_2SO_4. \:xH_2O has 50%50\% H2OH_2O. Hence, xx is :
    Solution
    The molar mass of Na2SO4Na_2SO_4 is 142142 g/mol.
    The molar mass of Na2SO4.xH2ONa_2SO_4.xH_2O is 142+18x142+18x g/mol. Since, 50%50\% water is present,
    142+18x2=18x\displaystyle \dfrac {142+18x}{2}=18x
    36x=142+18x36x=142+18x
    18x=14218x=142
    x=7.98x=7.9 \simeq 8
  • Question 5
    1 / -0
    Cortisone is a molecular substance containing 2121 atoms of carbon per molecule. The mass percentage of carbon in cortisone is 69.9869.98%. Its molar mass is:
    Solution
    Cortisone is a molecular substance containing 2121 atoms of carbon per molecule.
    11 mole of cortisone contains 21×12=25221 \times 12 = 252 g of CC.
    The mass percentage of carbon in cortisone is 69.9869.98%.
    Its molar mass will be 10069.98×252=360.1 \dfrac {100}{69.98} \times 252 = 360.1 g/mol.
  • Question 6
    1 / -0
    The sodium salt of methyl orange has 7%7\% sodium. What is the minimum molecular weight of the compound?
    Solution
    Let MM be the minimum molecular weight of the compound.

    7%7\% of the compound is sodium. It is equal to 7100×M=0.07M\dfrac {7}{100} \times M = 0.07M g sodium in 11 mole of compound.

    At least, 11 mole of NaNa will be present in 11 mole of compound.
    11 mole of sodium weighs 2323 g.

    Thus, 0.07M=230.07M=23

    M=230.07=329M=\dfrac {23}{0.07}=329 g/mol

    Thus, the minimum molecular weight of the compound is 329329 g/mol.

    Hence, the correct option is CC
  • Question 7
    1 / -0
    Mole fraction of ethanol and water mixture is 0.250.25. Hence, percentage concentration of ethanol by weight of mixture is:
    Solution

    Explanation:

    Molar mass of ethanol = 46 grams/(mole)46\ grams/(mole)

    Molar mass of water = 18 gram/(mole)18\ gram/(mole)

    The sum of all mole fractions is 11.

    The mole fraction of:

    Ethanol = 0.250.25 and water = 0.750.75

    Let the 10 moles of mixture are present.

    Moles of ethanol =0.2510×100=\dfrac {0.25}{10}\times 100 = 2.52.5

    Moles of water =0.7510×100=\dfrac {0.75}{10}\times 100= 7.57.5        

    Given mass of ethanol =2.5×46=115 grams=2.5 \times 46 = 115\ grams

    Given mass of water 7.5×18=135 grams7.5 \times 18 = 135\ grams

    The percentage concentration of ethanol by weight in the mixture =115115+135×100=\dfrac {115}{115+135}\times 100

    The percentage concentration of ethanol by weight in the mixture =46%=46\%.

    Hence, the correct answer is option CC.

  • Question 8
    1 / -0
    A spherical ball of radius 77 cm contains 56%56\% iron. If density is 1.41.4 g/cc, the number of moles of FeFe present approximately is :
    Solution
    The volume of the ball is 43πr3=43×3.1416×73=1436.7\dfrac {4}{3}\pi r^3 = \dfrac {4}{3} \times 3.1416 \times 7^3 = 1436.7 cc.
    The density is 1.41.4 g/cc.
    The mass of the ball is 1.4×1436.7=2011.51.4 \times 1436.7 = 2011.5 g.
    However, the percentage of iron is 56%56\%.
    The mass of iron is 56100×2011.5=1126.4\dfrac {56}{100}\times 2011.5 =1126.4 g.
    The molar mass of FeFe is 5656 g/mol.
    The number of moles of Fe present is 1126.456=20\dfrac {1126.4}{56}=20.
    Thus, the ball contains 2020 moles of FeFe.
  • Question 9
    1 / -0
    Which of the following is a suitable example for illustrating the law of conservation of mass?
     (Atomic mass of O = 16 g/mol, H = 1 g/mole)
    Solution
    The option (A) is a suitable example for illustrating the law of conservation of mass.
     
    18g of water is formed by the combination of 16g oxygen with 2g of hydrogen.

    Total mass of reactants (O and H) = 16 + 2 = 18 g.
    The mass of product (water) = 18 g.

    Since, the mass of product is equal to the total mass of reactants, the mass is conserved during the reaction, thus illustrating the law of conservation of mass.

    Note: The option (A) represents chemical change while all other options represent physical change. The law of conservation of mass corresponds to a chemical change.
  • Question 10
    1 / -0
    A salt is formed due to the reaction between an oxy acid containing chlorine and a base containing a monovalent metal of atomic mass xx. The number of oxygen atoms in one molecule of the acid is more than the corresponding 'ic' acid. Calculate the molecular mass of the salt.
    Solution
    An oxy-acid containing chlorine with the name ending with 'ic' acid is chloric acid, HClO3HClO_3.

    The number of oxygen atoms in one molecule of the acid is more than the corresponding 'ic' acid.

    Hence, our oxyacid is perchloric acid HClO4HClO_4.

    Let our base be MOHMOH where M is a monovalent metal of atomic mass xx.

    The acid base reaction is :

    HClO4+MOHMClO4+H2OHClO_4 + MOH \rightarrow MClO_4 + H_2O

    The atomic masses of metal, ClCl and OO are x g/mol,35.5g/mol x \ g/mol, 35.5 g/mol and 16g/mol16 g/mol respectively.

    The molecular mass of the salt MClO4MClO_4 is x+35.5+4(16)=99.5+xx + 35.5 + 4 (16) = 99.5 + x

    Hence, the correct answer is option B\text{B}.
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