Question

What is the molarity of a solution in which 7.1 g of sodium sulfate is dissolved in enough water to make 100. mL of solution?

Answer 1

Answer:

0.5 M

Explanation:

Hello,

In this case, since sodium sulfate molar mass is 142.04  g/mol we first need to compute the moles in 7.1 g as shown below:

$n=7.1g*\frac{1mol}{142.04 g} =0.050mol$

Then, since 100 mL are 0.1 L the molarity turns out:

$M=\frac{n}{V}=\frac{0.05mol}{0.1L}\\\\M=0.5M$

Regards.

Answer 2

Answer:

0.50 M

Explanation:

Given data

• Mass of sodium sulfate (solute): 7.1 g

• Volume of solution: 100 mL

Step 1: Calculate the moles of the solute

The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:

$7.1g \times \frac{1mol}{142.02g} = 0.050mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1 L = 1000 mL.

$100mL \times \frac{1L}{1000mL} =0.100L$

Step 3: Calculate the molarity of the solution

$M = \frac{moles\ of\ solute }{liters\ of\ solution} = \frac{0.050mol}{0.100L} =0.50 M$