Question

ANSWER THE 3 FOLLOWING QUESTIONS WITH EVIDENCE (40 POINTS):

Answer 1

Answer:

1.  3.02 liters

2. 302.2 liters

3. 2.83 grams

Explanation:

Question 1.

Let's convert the molarity in mass.

Molarity is mol/L, so if we convert the moles to mass, we are talking about a mass, in 1L of solution. (mol . molar mass)

0.5 mol . 110.98 g/m = 55.49 g

Now we can make a rule of three:

If 55.49 grams of chloride are in 1L of solution

then 167.5 g of chloride are in (167.5g . 1L) / 55.49 g = 3.02L

For question 2 and 3, we assume STP, where pressure is 1 atm and T° is 273K. - Standard conditions of temperature and pressure (STP)

Let's apply the Ideal gases law equation

P . V = n . R .T

1 atm . V = 13.5 moles . 0.082 l.atm / mol .K . 273K

V = (13.5 moles . 0.082 l.atm / mol .K . 273K) / 1 atm = 302.2 liters (question 2)

Question 3.

1 atm . 31.7 L = n . 0.082 l.atm / mol.K . 273K

(1 atm . 31.7 L ) / (0.082 l.atm / mol.K . 273K) = 1.41 moles → n

Let's convert moles to mass ( mol . molar mass)

1.41 mol . 2g/mol = 2.83 grams