# Determine the mass of a solution of ethanol in water containing 12.04 × 1022 carbon atoms and 24.08

**Determine the mass of a solution of ethanol in water containing 12.04 × 1022 carbon atoms and 24.08 × 1022 oxygen atoms.**

Given:

N (C) = 12.04 * 10 ^ 22

N total (O) = 24.08 * 10 ^ 22

To find:

m solution (С2Н5OH) -?

Decision:

1) Find the amount of substance C:

n (C) = N (C) / NA = 12.04 * 10 ^ 22 / 6.02 * 10 ^ 23 = 0.2 mol;

2) Find the amount of substance С2Н5OH (by definition):

n (C2H5OH) = n (C) / 2 = 0.2 / 2 = 0.1 mol;

3) Find the mass of the substance С2Н5OH:

m (C2H5OH) = n (C2H5OH) * Mr (C2H5OH) = 0.1 * 46 = 4.6 g;

4) Find the total amount of substance O:

n total (O) = N total (O) / NA = 24.08 * 10 ^ 22 / 6.02 * 10 ^ 23 = 0.4 mol;

5) Find the amount of substance O in С2Н5OH (by definition):

n (O in C2H5OH) = n (C2H5OH) = 0.1 mol;

6) Find the amount of substance O in H2O:

n (O in H2O) = n total (O) – n (O in C2H5OH) = 0.4 – 0.1 = 0.3 mol;

7) Find the amount of substance H2O (by definition):

n (H2O) = n (O in H2O) = 0.3 mol;

8) Find the mass of the substance H2O:

m (H2O) = n (H2O) * Mr (H2O) = 0.3 * 18 = 5.4 g;

9) Find the mass of the C2H5OH solution:

m solution (C2H5OH) = m (C2H5OH) + m (H2O) = 4.6 + 5.4 = 10 g.

Answer: The mass of the C2H5OH solution is 10 g.