34.2 gm of sucrose (C12 H22 O11) are dissolved in 180 gm of water calculated the number of oxygen atoms in the solution
Answers
Given,
=> 34.2 grams of Sucrose (C12H22O11) is dissolved in 180 grams of water.
To find,
=> Number of oxygen atoms in the solution.
Concepts Involved,
=> 1 mole of any compound contains 6.022*10^23 molecules of it.
We know that,
=> Molar mass of Sucrose (C12H22O11) = 12*12+22*1+11*16
=> 342 grams per mole.
=> Molar mass of water (H2O) = 2*1+16 => 18 grams per mole.
Hence,
=> 34.2 grams of Sucrose is 34.2/342 moles => 0.1 mol of C12H22O11
=> 180 grams of H2O is 180/18 moles => 10 mol of H2O
∵ 1 mole of a compound has 6.022*10^23 molecules in it.
∴ 0.1 mol of C12H22O11 has 6.022*10^23*0.1 molecules of sucrose.
=> 6.022*10^22 molecules of sucrose are present.
∴ 10 mol of H2O has 6.022*10^23*10 molecules of water.
=> 6.022*10^24 molecules of water are present.
Now,
∵ One molecule of water (H2O) has 1 atom of Oxygen.
∵ One molecule of Sucrose (C12H22O11) has 11 atoms of Oxygen.
∴ 6.022*10^24 molecules of H2O has 6.022*10^24*1 atoms of O.
=> 6.022*10^24 atoms of Oxygen in 180 grams of Water.
∴ 6.022*10^22 molecules of C12H22O11 has 6.022*10^22*11 atoms of O.
=> 66.242*10^22 atoms of Oxygen in 34.2 grams of Sucrose.
=> Total number of Oxygen atoms present:
=> 6.022*10^24 + 66.242*10^22 atoms of Oxygen.
=> 10^22 ( 6.022*10^2 + 66.242) atoms of Oxygen.
=> 10^22 ( 602.2 + 66.242)
=> 10^22 (668.442)
=> 6.68442 * 10^2 * 10^22
=> 6.68442 * 10^24 atoms of Oxygen.
=>Hence, the solution will have 6.68442 * 10^24 atoms of Oxygen in it.