Question

How many h atoms are in 3.4 g of c12h22o11?

Answer 1

The right answer is 45

Answer 2

Answer:

The hydrogen atom's number in $3.4g$ of $C_{12} H_{22} O_{11}$ = $1.31\times 10^{23}$ atoms.

Explanation:

Data given,

The mass of $C_{12} H_{22} O_{11}$ = $3.4g$

The hydrogen atom's number in $3.4g$ $C_{12} H_{22} O_{11}$ =?

Firstly, we have to calculate the molar mass of $C_{12} H_{22} O_{11}$:

• The molar mass of $C_{12} H_{22} O_{11}$ = ( $12$ × C-atom's atomic mass ) + ( $22$ × H-atom's atomic mass ) + ( $11$ × O-atom's atomic mass )
• The molar mass of $C_{12} H_{22} O_{11}$ = $(12\times 12) + (22\times 1) + ( 11 \times 16 )$ = $144+ 22+176$ = $342g/mol$

Now,

• The number of moles of $C_{12} H_{22} O_{11}$ = $\frac{Given mass}{Molar mass}$ = $\frac{3.4}{342}$ = $9.94\times 10^{-3} mol$

As we know,

• $1mole$ = $6.022\times 10^{23}$ molecules
• $9.94\times 10^{-3} mol$ = $9.94\times 10^{-3} \times 6.022\times 10^{23}$ molecules = $5.98\times 10^{21}$ molecules

In one molecule of $C_{12} H_{22} O_{11}$, $22$ H-atoms are present.

Therefore,

• The hydrogen atom's number in $3.4g$ $C_{12} H_{22} O_{11}$ = $22\times 5.98\times 10^{21}$ atoms = $1.31\times 10^{23}$ atoms.