Question

please ans the question​

Answer 1

Answer:

1. 342.30 g/mol

2.There are 12 Carbon atoms, 22 Hydrogen atoms and 11 Oxygen atoms in one molecule of sucrose. Therefore, in 6.022 × 10^23 molecules of sucrose, there are

=(12+22+11)∗(6.022×1023)  

=(45)∗(6.022×1023)  

=270.99×1023  atoms!

Explanation:

MARK AS BRAINIEST

Answer 2

Answer:

Compound given :-

Sucrose = $C_{12} H_{22} O_{11}$

i) Molecular Formula

Step 1

multiply all the elemental atomic weights to their respective suffixes that are present in the molecular formula ($C_{12} H_{22} O_{11}$)

• C = 12g x 12 = 144g
• H = 1g x 22 = 22g
• 0 = 16g x 11 = 176g

(Note : the atomic weight of C = 12g ; H = 1g ; O= 16g)

Step 2

Add the values obtained in step 1 to get the molecular formula

= 144g + 22g + 176g

= 342g

The molecular formula of Sucrose ($C_{12} H_{22} O_{11}$) is 342g. (Ans

-----------------------------------------------------------------------------------------------------------

ii) Number of atoms in 1 molecule of sucrose

The number of atoms of each element in a molecule is given by the suffix present in its molecular formula.

(For eg, In H2O number of H atoms = 2 and no. of O atoms = 1)

=> Calculating the number of atoms of each type :-

• No. of Carbon(c) atoms = 12

• No. of Hydrogen(H) atoms = 22

• No. of Oxygen(O) atoms = 11

=> Total number of atoms in 1 molecule of Sucrose = 12 + 22 + 11

= 12 + 33

= 45

The no. of atoms present in 1 molecule of Sucrose ($C_{12} H_{22} O_{11}$) is 45. (Ans