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Chemistry Test - 5

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Chemistry Test - 5
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  • Question 1
    1 / -0.25

    The monomers used in the preparation of dextron are

    Solution

    The monomers used in the preparation of dextron are lactic acid and glycolic acid. There action for the formation of dexron can be written as

  • Question 2
    1 / -0.25

    In which oxidation state, group 15 elements act as Lewis base?

    Solution

    Group 15 elements acts as Lewis base in - 3 oxidation state due to availability of lone pair of electrons on the central atom. This basic character of group 15 decreases down the group with an increase in the size of the central atom.

  • Question 3
    1 / -0.25

    Which of the following elements does not react with hot concentrated sulphuric acid?

    Solution

    Nitrogen doesn’t react with concentrated sulphuric acid as its bond dissociating energy is highest among the given elements. This is because of its small size.

  • Question 4
    1 / -0.25

    The tonic charges of m anganate and permanganate ion are respectively

    Solution

    Chemical formula for manganate and permanganate ion respectively are MnO2-4 and MnO-4 . Thus, the ionic charge of manganate and permanganate ion are respectively -2 and -1.

  • Question 5
    1 / -0.25

    The enzyme which converts maltose to glucose is

    Solution

    The enzyme which converts maltose to glucose is maltase. During digestion, starch is partially transformed into maltose by pancreatic or salivary enzymes called amylases; maltase secreted by the intestine men converts maltose into glucose. The glucose so produced is either utilised by the body or stored in the liver as glycogen.

  • Question 6
    1 / -0.25

    What is the atomicity of aluminium phosphate?

    Solution

    The chemical formula for aluminium phosphate is AIPO4 .

    Atomicity of heteronuclear molecule = total number of atoms present

    ∴ Atomicity of AlPO4 = 1 +1 + 4 = 6

  • Question 7
    1 / -0.25

    Standard hydrogen electrode (SHE) is a

    Solution

    Standard hydrogen electrode (SHE) is the primary reference electrode. SHE is represented by Pt(s)|B2 (g)|H+ (ag), is assigned a zero potential at all temperatures corresponding to reaction.

  • Question 8
    1 / -0.25

    Which of the following sets of solutions of urea (mol, mass 60 g mol-1) and sucrose (mol. mass 342 g mol-1) is isotonic?

    Solution

    Key idea: Isotonic solutions are those solutions which have the same osmotic pressure at a given temperature.

    Formula for osmotic pressure, π = CRT

    considering the set given in option (d), i.e. 30 gL-1 urea and 17.1 gL-1 sucrose.

    Given, the molecular mass of urea 60 g mol-1 and molecular mass of sucrose 342 g mol-1 .

    For urea.

    conc, C = 3/60 = 1/20

    Osmotic pressure π1 = (1/20) x R x T

    For sucrose conc, C = 17.1/342 = 1/20

    ∴ Osmotic pressure, π2 = (1/20) RT

    Thus, the set of solutions of urea and sucrose given in option (d) is isotonic.

  • Question 9
    1 / -0.25

    Which of the following is a natural polymer?

    Solution

    Natural polymers are those polymers which are found in plants and animals. Among the given polymers liner is a natural polymer which is made from cellulose (polymer of glucose molecules).

  • Question 10
    1 / -0.25

    When a mixture of manganese dioxide, potassium hydroxide and potassium chlorate is fused, the product obtained is

    Solution

    When a mixture of manganese dioxide, potassium hydroxide and potassium chlorate is fused then potassium chlorate first decomposes to give potassium chloride and oxygen gas. The formed O2 gas then reacts with MnO2 and KOH to give K2 MnO4 . The equations for the above reaction can be written as

    2MnO2 + 4KOH + O2 → 2K2 MnO2 + 2H2 O

  • Question 11
    1 / -0.25

    Relationship between van't Hoff's factor (i) and degree of dissociation (α) is

    Solution

    Relationship between van't Hoff factor(i) and degree of dissociation (α) is given by

    α = 1-i/n'-1

    where, n is the number of ions formed alter dissociation.

    The relationship can be obtained as follows;

    For the reaction, A ⇌ n'B

    Initially 1 mole 0

    After dissociation (1 - α) mole n' α

    Total number of moles present in the solution

    = (1 - α) + n'α = 1 + (n'-1)α

    van't Hoff factor, i = 1 + (n' - 1), α > 1 if n' ≥ 2

    ∴ α = i-1/n'-1

  • Question 12
    1 / -0.25

    In the reaction iodide ton acts as

    Solution

    For there action,

    Iodide ion acts a homogenous catalyst as the phase of I- and reactant, H2 O2 is aqueous (same).

  • Question 13
    1 / -0.25

    How many gram of sodium (atomic mass 23 u) is required to prepare one mole of ethane from methyl chloride by Wurtz reaction?

    Solution

    The Wortz reaction between methyl chloride and sodium metal in presence of dry ether can be written as

    ⇒ 2 moles of sodium metal reacts to give 1 mole of ethane.

    Also,

    1 mole of Na metal = 23 g

    ∴ 2 motesol Na metal = 23x 2 = 46g

    Thus, 46 g of sodium is required to prepare one mole of ethane from methyl chloride by Wurtz reaction.

  • Question 14
    1 / -0.25

    If C(s) + O2(g) → CO2(g), ΔH = -X,

    CO(g) + (1/2)O2 (g) → CO2 (g), ΔH = -Y,

    Calculate Δr H for CO(g) formation

    Solution

    C(s) + O2 (g) → CO2 (g), ΔH1 = -X, …(i)

    CO(g) + (1/2)O2 (g) → CO2 (g), ΔH2 = -Y, ….(ii)

    For the formation of CO subtract Eqs. (ii) from (i), i.e,

    ∴ Δr H for formation of CO = ΔH1 - ΔH2

    = -x+y or y-x

  • Question 15
    1 / -0.25

    Which among the following compounds is obtained when ethane nitrile is acid  hydrolysed?

    Solution

    Acetic acid is obtained when ethanenitrile is acid hydrolysed. The equation for the reaction can be written as

  • Question 16
    1 / -0.25

    9 gram anhydrous oxalic acid (mol. wt. = 90) was dissolved in 9.9 moles of water. If vapour pressure of pure wafer is pº1the vapour pressure of solution is

    Solution

    The total vapour pressure of a solution in this case only depends on vapour pressure of water as anhydrous oxalic acid is a non-vola tile compound.

    ∴ Vapour pressure of solution = vapour pressure of water (pw )

    According to Raoutt’s law pw = xww

    Number of moles of oxalic acid = 9/90 = 0.1 moles

    ∴ xw = 9.9/9.9+0.1 = 0.99

    ⇒ ps = pw = 0.99 x pº1

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