In laboratory experiment, a NOVDEC Student was required to prepare 500 cm3 of Im Solution of
glucose (c6, H12,06) Determine the
i Molar
Mass
ii) Amount of ghicoseB. In in moles in the Solrition
[ C= 12, H = 10, 0=16]

i. Molar mass of glucose = 180 g/mol

ii. Amount of glucose = 0.5 mole

Explanation:

The volume of the glucose solution to be prepared = 500

Molarity of the glucose solution to be prepared = 1 M

i. Molar mass of glucose () = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol

ii. mole = molarity x volume. Hence;

amount (in moles) of the glucose solution to be prepared

                 = 1 x 500/1000 = 0.5 mole

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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