Question

One mole of any substance contains $6.022 \times 10^{23}$atoms/molecules. Number of molecules of $H_{2}SO_{4}$present in 100 mL of $0.02MH_{2}SO_{4}$solution is_____ molecules.

• A. $12.044 \times 10^{20}$
• B. $6.022 \times 10^{23}$
• C. $1 \times 10^{23}$
• D. $12.044 \times 10^{23}$

Answer 1

Answer: (A)

$12.044 \times 10^{20}$

Answer 2

(1) : Molarity

$= \frac{No.ofmolesofsolute \times 1000}{Volumeofsolution(inmL)}$

$\because$1M $H_{2}SO_{4}$contains 1 mole of $H_{2}SO_{4}$in 1000 mL of solution

$\therefore 0.02MH_{2}SO_{4}$in 100 mL of solution contains

$= \frac{0.02 \times 100}{1000} = 2 \times 10^{- 3}mol$

$= 2 \times 10^{- 3} \times 6.022 \times 10^{23}$

$= 12.044 \times 10^{20}molecules`$