Question

Question 9
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
(a) $6.68\times {10}^{23}$
(b) $6.09\times {10}^{22}$
(c) $6.022\times {10}^{23}$
(d) $6.022\times {10}^{21}$

Answer 1

Sucrose molar mass = 144 + 22 + 176 = 342 g/mol

1 mol of sucrose contains = 11× $N_A$ atoms of oxygen,

where $N_A$ = 6.023×${10}^{23}$

0.01 mol of sucrose ($C_{12}{H}_{22}{O}_{11}$) contains = 0.01 × 11 ×$N_A$ atoms of oxygen

= 0.11× $N_A$ atoms of oxygen

= 18 g/(1 × 2+ 16)g$mo{l}^{-1}$

=18 g /18 g$mo{l}^{-1}$

= 1mol

1mol of water (H2OH_2O$H_{2}O$) contains 1 × $N_A$ atom of oxygen

Total number of oxygen atoms = Number of oxygen atoms from sucrose + Number of oxygen atoms from water

Total number of oxygen atoms = 0.11 $N_A$ + 1.0 $N_A$ = 1.11$N_A$

Number of oxygen atoms in solution = 1.11 × Avogadro’s number

Number of oxygen atoms in solution = 1.11 × 6.022 ×${10}^{23}$

Number of oxygen atoms in solution = 6.68 × ${10}^{23}$