Question

solution of glucose C6H12O6 (Molar mass.
holecules. The concentration of the
50 mL of an aqueous solution of glucose C6H12
180 g/mol) contains 6.02 x 1022 molecules. The concess
solution will be
(A) 01 M
(B) 0-2 M
(C) 1.0 M
(D) 2.0 M​

Answer 1

Answer:

No. Of miles of sucrose is =Given particles/Avagadro number

=6.022 x 10^22/6.022 x 10^23

=10^(-1)

=0.1

Molarity

= no. Of moles of solute/Volume in liters

=0.1/(50/1000)

=2.0 M

Answer is (D)-->2.0M

P.S..mai exam mein galat karke aaya hoon....pr ab sahi ho gaya...kya faida

Explanation:

Answer 2

The Concentration of the solution will be : D) 2.0 M

Given,

• Volume of the solution = 50 mL
• Number of molecules of Glucose in given solution = 6.02 * 10^22
• We know,
• Number of molecules in 1 mole of any element = 6.023 * 10^23
• Therfore,  number of moles of Glucose in given solution =                        6.023 * 10^22 / 6.023 * 10^23          =   0.1 moles
• Now, we know Concentration of a solution = number of moles of solute in 1 L of solution
•      M = no. of moles * 1000 / Volume in ml.
•      M = 0.1 * 1000 / 50   =  2
• Therfore, Concentration of solution = 2 M.