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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?

Respuesta :

dsdrajlin

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:

[tex]7.1g \times \frac{1mol}{142.02g} = 0.050mol[/tex]

Step 2: Convert the volume of solution to liters

We will use the relation 1 L = 1000 mL.

[tex]100mL \times \frac{1L}{1000mL} =0.100L[/tex]

Step 3: Calculate the molarity of the solution

[tex]M = \frac{moles\ of\ solute }{liters\ of\ solution} = \frac{0.050mol}{0.100L} =0.50 M[/tex]

sebassandin

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:

[tex]n=7.1g*\frac{1mol}{142.04 g} =0.050mol[/tex]

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

[tex]M=\frac{n}{V}=\frac{0.05mol}{0.1L}\\\\M=0.5M[/tex]

Regards.

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