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

2020 · 3 Sep · Shift 2 · Q18
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. /2020 · 3 Sep · Shift 2 · Q18

Some Basic Concepts of Chemistry question

2020 · 3 Sep · Shift 2 · Q18

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
6.023 ×\times× 1022 molecules are present in 10 g of a substance 'x'. The molarity of a solution containing 5 g of substance 'x' in 2 L solution is ‾\underline{\hspace{2cm}}​ × 10-3
Numerical answer
View written solutionFree

Correct answer: 25

  1. Find the number of moles in 10 g of substance xxx

Given number of molecules: 6.023×10226.023 \times 10^{22}6.023×1022

We know Avogadro's number is: NA=6.023×1023N_A = 6.023 \times 10^{23}NA​=6.023×1023

So, moles in 10 g are: n=6.023×10226.023×1023=10−1=0.1 moln = \frac{6.023 \times 10^{22}}{6.023 \times 10^{23}} = 10^{-1} = 0.1 \text{ mol}n=6.023×10236.023×1022​=10−1=0.1 mol

  1. Find the molar mass of substance xxx

If 10 g corresponds to 0.10.10.1 mol, then molar mass is: M=100.1=100 g mol−1M = \frac{10}{0.1} = 100 \text{ g mol}^{-1}M=0.110​=100 g mol−1

  1. Find moles in 5 g of substance xxx

n=5100=0.05 moln = \frac{5}{100} = 0.05 \text{ mol}n=1005​=0.05 mol

  1. Find molarity of the solution

Volume of solution = 2 L

Molarity=0.052=0.025 mol L−1\text{Molarity} = \frac{0.05}{2} = 0.025 \text{ mol L}^{-1}Molarity=20.05​=0.025 mol L−1

Now write this in the form _×10−3\_ \times 10^{-3}_×10−3: 0.025=25×10−30.025 = 25 \times 10^{-3}0.025=25×10−3

  1. Final Answer

The required number is: 25\boxed{25}25​

  1. Comparison with stored answer

Stored correct answer = 25

Our derived answer matches the stored answer.

PreviousNext

More from Some Basic Concepts of Chemistry

  • The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is ​.2020 · Numerical
  • A 20.0 mL solution containing 0.2 g impure H2​O2​ reacts completely with 0.316 g of KMnO4​ in acid solution. The purity of H2​O2​ (in %) is ​ (mol. wt. of H2​O2​ = 34; mol. wt. of KMnO4​ = 158)2020 · Numerical
  • The mass of ammonia in grams produced when 2.8 kg of dinitrogen quantitatively reacts with 1 kg of dihydrogen is ​.2020 · Numerical
  • Consider the following equations : 2Fe2+ + H2​O2​ → xA + yB (in basic medium) 2MnO4−​ + 6H+ + 5H2​O2​ → x'C + y'D + z'E (in acidic medium) The sum of the stoichiometric coefficients x, y, x', y', and z' for products…2020 · Numerical
  • A solution of two components containing n1​ moles of the 1st component and n2​ moles of the 2nd component is prepared. M1​ and M2​ are the molecular weights of component 1 and 2 respectively. If d is the density of the solution in…2020 · MCQ
  • The average molar mass of chlorine is 35.5 g mol–1. The ratio of 35Cl to 37Cl in naturally occuring chlorine is close to :2020 · MCQ
  • Amongst the following statements, that which was not proposed by Dalton was :2020 · MCQ
  • The flocculation value of HCl for arsenic sulphide sol. is 30 m mol L-1 If H2​SO4​ is used for the flocculatiopn of arsenic sulphide, the amount in grams, of H2​SO4​ in 250 ml required for the above purposed is ​…2020 · Numerical