Question

Calculate the number of hydrogen and oxygen molecules in a solution prepared by mixing 45g of glucose in 36g of water

Answer 1

1 molecules of glucose $C_{6}H_{12}O_{6}$ has 6 atoms of oxygen and 12  atoms of hydrogen

So, number of glucose molecules in 45 g

$\frac{45}{180}*6.023*10^{23} =  1.51*10^{23} molecules$

Thus , number of oxygen molecules = $\frac{6}{2}*1.51*10^{23} =   4.53*10^{23}$

number of hydrogen molecules = $\frac{12}{2}*1.51*10^{23}  = 9.06*10^{23}$

In case of water , there are 1 H and 2 O atoms or 1 molecule present in 1 molecule of water

Number of water molecules in 36 g

$= \frac{36}{18}*6.023*10^{23} =   12.046*10^{23}$ molecules

Number of Hydrogen molecules = $\frac{12.046*10^{23} }{2} = 6.023*10^{23}$

Number of Oxygen molecules = $12.046*10^{23}$

Therefore, total molecules of oxygen in mixture = $(4.53+12.046)*10^{23} = 16.576*10^{23}$

Total number of hydrogen molecules = $(9.06+6.023)*10^{23} =  15.083*10^{23}$

Answer 2

Answer: oxygen molecules - 1.052x 10^24

Hydrogen molecules - 2.1 x 10^24

Explanation: No. of molecules of gulcose in 45gm= 45/180*6.022*10^23 = 1.50 *10^23

No of molecules of O2 = 3* 1.50*10^23= 4.5*10^23

No of hydrogen molecules = 6*1.50*10^23 = 9*10^23

In H2O of 36GM.

No. Of molecules of H20 36/18*6.022*10^23

No. Of hydrogen molecules = 12.04 * 10^23

No. Of oxygen = 6.022*10^23

Total oxygen molecules = (4.5+6.022)*10^23= 1.052*10^24

Total hydrogen molecules=( 12.04+9)*10^23= 2.10*10^24