Solution 'X' contains Na2CO3 and NaHCO3, 20 mL of X when titrated

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381.

Solution 'X' contains Na2CO3 and NaHCO3, 20 mL of X when titrated using methyl orange indicator consumed 60 mL of 0.1 M HCl solution. In another experiment, 20 mL of X solution when titrated using phenolphthalein, consumed 20 mL of 0.1 M HCl solution. The concentrations (in mol L-1) of Na2CO3 and NaHCO3 in X are respectively

  • 0.01, 0.02

  • 0.1, 0.1

  • 0.01, 0.01

  • 0.1, 0.01


B.

0.1, 0.1

For a titration of a basic solution of Na2CO3 and NaHCO3 against HCl, if phenolphthalein is used as indicator, the end product is indicated only for half neutralization of Na2CO3i.e. upto NaHCO3

Na2CO+ HCl → NaHCO+ NaCl

The remaining solution then contains the unreacted  NaHCO3 from this reaction plus the unreacted NaHCO3 originally in the solution. At the phenolphthalein end point, there is no reaction between HCl and NaHCO3.

From the equation,

Mole of HCl consumed = mol of Na2CO3

         10 mol of 0.1 M = 20 mol of 0.1 M

 The concentration of Na2CO3 in solution X = 0.1 M

For a quantity of Na2COexactly half volume of the HCl is used at the phenolphthalein
end point and the second half volume of the HCl is required for complete neutralization of Na2CO3 at methyl orange end point.

NaHCO3 + HCl → NaCl + CO2 + H2O

 The volume of HCl required to neutralize Na2COin original sample = 2 × 20 mL = 40 mL

If methyl orange is used, the end point is indicated when all the alkali is neutralized.

As 40 mL of 0.1 M HCl is consumed in complete neutralization of Na2CO3 at methyl ornage end point from the original sample would be reamining HCl

= (60 - 40) mL 

= 20 mL

Now, 

1 mol of NaHCO3 = 1 mol of HCl

 0.1 mol of NaHCO3 = 0.1 mol of HCl


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382.

The number of moles of electrons required to deposit 36 g of Al from an aqueous solution of Al(NO3)3 is (At. wt. of Al = 27)

  • 4

  • 3

  • 2

  • 1


383.

A carbon compound contains 12.8% of carbon, 21.% of hydrogen and 85.1% of bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula of the compound.

(Atomic weight : H = 1.008; C = 12.0; Br = 79.9)

  • CH3Br

  • CH2Br2I

  • C2H4Br2

  • C2H3Br3


384.

3.011 × 1022 atoms of an element weight 1.15 gm. The atomic mass of the element is

  • 23

  • 10

  • 16

  • 35.5


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385.

A gas 'X' is dissolved in water at 2 bar pressure. Its mole fraction is 0.02 in solution. The mole fraction of water when the pressure of gas is doubled at the same temperature is

  • 0.04

  • 0.98

  • 0.96

  • 0.02


386.

KMnO4 reacts with KI in basic medium to form I2 and MnO2. When 250ml of 0.1M KI solution is mixed with 250ml of 0.02M KMnO4 in basic medium, what is the number of moles of I2 formed?

  • 0.015

  • 0.0075

  • 0.005

  • 0.01


387.

The oxide of a metal contains 40% of oxygen. The valency of metal is 2. What is the atomic weight of metal?

  • 24

  • 13

  • 40

  • 36


388.

Which of the following reactions does not take place?

  • F2 + 2Cl- → 2F- + Cl2

  • Br2 + 2I- → 2Br- + I2

  • Cl2 + 2Br- → 2Cl- + Br2

  • Cl2 + 2F- → 2Cl- + F2


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389.

What weight of hydrated oxalic acid should be added for complete neutralisation of 100 mL of 0.2 N -NaOH solution?

  • 0.45 gm

  • 0.90 gm

  • 1.08 gm

  • 1.26 gm


390.

The number of moles of a solute in its solution is 20 and total number of moles are 80. The mole fraction of solute is

  • 2.5

  • 0.25

  • 1

  • 0.75


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