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12. How many molecules of glucose, C6H12O6, are present in a 152 g sample
A.) How many atoms of carbon are present in the sample?
B.) How many atoms in total are present in the sample?
13. Calculate the mass in grams of 2.37 x 1025 formula units of aluminum

Respuesta :

transitionstate

Q.No. 12:

Answer:

                 5.08 × 10²³ Glucose Molecules

Solution:

Data Given:

                Mass of Glucose  =  152 g

                M.Mass of Glucose  =  180.156 g.mol⁻¹

Step 1: Calculate Moles of Glucose as,

                Moles  =  Mass ÷ M.Mass

Putting values,

                Moles  =  152 g ÷ 180.156 g.mol⁻¹

                Moles  =  0.8437 mol

Step 2: Calculate number of Glucose Molecules,

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of glucose molecules can be written as,

 Moles  =  Number of Glucose Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Glucose Molecules,

 Number of Glucose Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

 Number of Glucose Molecules  =  0.8437 mol × 6.022 × 10²³ Atoms.mol⁻¹

 Number of Glucose Molecules =  5.08 × 10²³ Glucose Molecules

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Q.No. 12: (A)

Answer:

                3.04 × 10²⁴ Carbon Atoms

Solution:

              The molecular formula of Glucose is C₆H₁₂O₆. This specifies that there are six carbon atoms in one molecule of Glucose.

Hence,  when,

               1 molecule of Glucose contain  =  6 atoms of Carbon

Then,

     5.08 × 10²³ Glucose Molecules will contain  =  X atoms of Carbon

Solving for X,

                     X =  5.08 × 10²³ molecules  × 6 atoms / 1 molecule

                     X  =  3.04 × 10²⁴ Carbon Atoms

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Q.No. 12: (B)

Answer:

                1.22 × 10²⁵ Atoms in total

Solution:

              The molecular formula of Glucose is C₆H₁₂O₆. This specifies that there are 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms in one molecule of Glucose. So, there are 24 atoms in one molecule of glucose

Hence,  when,

                    1 molecule of Glucose contain  =  24 atoms

Then,

          5.08 × 10²³ Glucose Molecules will contain  =  X atoms

Solving for X,

                     X =  5.08 × 10²³ molecules  × 24 atoms / 1 molecule

                     X  =  1.22 × 10²⁵ Atoms in total

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Q. No. 13

Answer:

                   1061.81 g of Aluminium

Solution:

Data Given:

                Number of F.Units  =  2.37 × 10²⁵

                A.Mass of Aluminium  =  26.98 g.mol⁻¹

                Mass of Aluminium  =  ?

Step 1: Calculate Moles of Aluminium,

                  Moles  =  Number of F.Units ÷ 6.022 × 10²³ F.Units.mol⁻¹

Putting value,

                  Moles  =   2.37 × 10²⁵ F.Units ÷ 6.022 × 10²³ F.Units.mol⁻¹

                  Moles  =  39.35 mol

Step 2: Calculate Mass of Aluminium as:

                  Moles  =  Mass ÷ A.Mass

Solving for Mass,

                  Mass  =  Moles × A.Mass

Putting values,

                  Mass  =  39.35 mol × 26.98 g.mol⁻¹

                  Mass  =  1061.81 g of Aluminium

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