Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Some Basic Concepts of Chemistry question

2021 · 16 Mar · Shift 2 · Q4
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Some Basic Concepts of Chemistry
  5. /2021 · 16 Mar · Shift 2 · Q4

Some Basic Concepts of Chemistry question

2021 · 16 Mar · Shift 2 · Q4

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
The exact volumes of 1 M NaOHNaOHNaOH solution required to neutralise 50 mL of 1 M H3PO3H_3PO_3H3​PO3​ solution and 100 mL of 2 M H3PO2H_3PO_2H3​PO2​ solution, respectively, are :
  1. A
    100 mL and 50 mL
  2. B
    100 mL and 200 mL
  3. C
    100 mL and 100 mL
  4. D
    50 mL and 50 mL
View written solutionFree

Correct answer: B

  1. Identify the basicity of the acids

    For oxyacids of phosphorus, only the hydrogen atoms attached to oxygen are ionisable.

    • H3PO3H_3PO_3H3​PO3​ has structure HPO(OH)2HPO(OH)_2HPO(OH)2​. So it has 2 ionisable H atoms ⇒\Rightarrow⇒ it is dibasic.

    • H3PO2H_3PO_2H3​PO2​ has structure H2PO(OH)H_2PO(OH)H2​PO(OH). So it has 1 ionisable H atom ⇒\Rightarrow⇒ it is monobasic.

  2. For 50 mL50\,\text{mL}50mL of 1 M  H3PO31\,M\; H_3PO_31MH3​PO3​

    Volume =50 mL=0.050 L= 50\,\text{mL} = 0.050\,L=50mL=0.050L

    Moles of H3PO3H_3PO_3H3​PO3​: n=M×V=1×0.050=0.050 moln = M \times V = 1 \times 0.050 = 0.050\,\text{mol}n=M×V=1×0.050=0.050mol

    Since H3PO3H_3PO_3H3​PO3​ is dibasic, moles of NaOHNaOHNaOH required: 0.050×2=0.100 mol0.050 \times 2 = 0.100\,\text{mol}0.050×2=0.100mol

    For 1 M  NaOH1\,M\; NaOH1MNaOH, volume required is: V=nM=0.1001=0.100 L=100 mLV = \frac{n}{M} = \frac{0.100}{1} = 0.100\,L = 100\,\text{mL}V=Mn​=10.100​=0.100L=100mL

  3. For 100 mL100\,\text{mL}100mL of 2 M  H3PO22\,M\; H_3PO_22MH3​PO2​

    Volume =100 mL=0.100 L= 100\,\text{mL} = 0.100\,L=100mL=0.100L

    Moles of H3PO2H_3PO_2H3​PO2​: n=M×V=2×0.100=0.200 moln = M \times V = 2 \times 0.100 = 0.200\,\text{mol}n=M×V=2×0.100=0.200mol

    Since H3PO2H_3PO_2H3​PO2​ is monobasic, moles of NaOHNaOHNaOH required: 0.200×1=0.200 mol0.200 \times 1 = 0.200\,\text{mol}0.200×1=0.200mol

    For 1 M  NaOH1\,M\; NaOH1MNaOH, volume required is: V=nM=0.2001=0.200 L=200 mLV = \frac{n}{M} = \frac{0.200}{1} = 0.200\,L = 200\,\text{mL}V=Mn​=10.200​=0.200L=200mL

  4. Final result

    Required volumes of 1 M  NaOH1\,M\; NaOH1MNaOH are: 100 mL and 200 mL100\,\text{mL} \text{ and } 200\,\text{mL}100mL and 200mL

  5. Option check

    • A: 100 mL100\,\text{mL}100mL and 50 mL50\,\text{mL}50mL →\rightarrow→ incorrect
    • B: 100 mL100\,\text{mL}100mL and 200 mL200\,\text{mL}200mL →\rightarrow→ correct
    • C: 100 mL100\,\text{mL}100mL and 100 mL100\,\text{mL}100mL →\rightarrow→ incorrect
    • D: 50 mL50\,\text{mL}50mL and 50 mL50\,\text{mL}50mL →\rightarrow→ incorrect

Therefore, the correct answer is B.

PreviousNext

More from Some Basic Concepts of Chemistry

  • 15 mL of aqueous solution of Fe2+ in acidic medium completely reacted with 20 mL of 0.03 M aqueous Cr2​O 72−​. The molarity of the Fe2+ solution is ​× 10-2 M. (Round off to the Nearest…2021 · Numerical
  • The mole fraction of a solute in a 100 molal aqueous solution is ​× 10 − 2. (Round off to the Nearest Integer). [Given : Atomic masses : H : 1.0 u, O : 16.0 u ]2021 · Numerical
  • Consider the above reaction. The percentage yield of amide product is ​. (Round off to the Nearest Integer). (Given : Atomic mass : C : 12.0 u, H : 1.0 u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u) Includes diagram2021 · Numerical
  • ​ grams of 3-Hydroxy propanal (MW = 74) must be dehydrated to produce 7.8 g of acrolein (MW = 56) (C3​H4​O) if the percentage yield is 64. (Round off to the Nearest Integer). [Given : Atomic masses : C : 12.0 u,…2021 · Numerical
  • A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n × 10 − 1, when n = ​. (Round off…2021 · Numerical
  • Complete combustion of 3g of ethane gives x × 1022 molecules of water. The value of x is ​. (Round off to the Nearest Integer). [Use : NA = 6.023 × 1023; Atomic masses in u : C : 12.0; O : 16.0; H :…2021 · Numerical
  • 10.0 mL of Na2​CO3​ solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings : 4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL Based on these readings, and convention of titrimetric estimation…2021 · Numerical
  • 250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ​× 1021. (Nearest integer) (NA = 6.022 × 1023)2021 · Numerical