Respuesta :

mppalenciaa

Answer: 0.94L

Explanation:

As a solution of concentration of 0.32 M indicates that contains 0.32 moles of the compound in 1L of solution. So we need to calculate the grams of Mg(NO3)2 in 0.32 M:

[tex]\frac{0.32 moles Mg(NO3)2}{L}[/tex]×[tex]\frac{148.3 g Mg(NO3)2}{molMg(NO3)2}[/tex][tex]=\frac{47.5 g Mg(NO3)2}{L}[/tex]

This is telling you that you have 47.5 g de Mg(NO3)2 per liter of solution.

Now, using a rule of three you can know de volume that occupies a solution containing 45 g of Mg(NO3)2:

47.5 g Mg(NO3)2 → 1L solution

45 g Mg(NO3)2 → x

[tex]x=\frac{45g * 1L}{47.5 g} = 0.94L[/tex]

Finally, we have 45g of Mg(NO3)2 in a volume of 0.94L of solution 0.32 M Mg(NO3)2.

Lanuel

The volume in liters (L) of [tex]Mg(NO_3)_2[/tex] contained in 0.32 M is 0.95 grams.

Given the following data:

  • Number of moles = 0.32 M
  • Mass = 45 g

Molar mass of [tex]Mg(NO_3)_2[/tex] = 148.3 g/mol.

To calculate the volume in liters (L) of the given chemical solution:

First of all, we would determine the mass of [tex]Mg(NO_3)_2[/tex] contained in 0.32 M:

[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 0.32 \times 148.3[/tex]

Mass of [tex]Mg(NO_3)_2[/tex] = 47.46 grams

Now, we can calculate the volume in liters (L);

47.46 grams of [tex]Mg(NO_3)_2[/tex] = 1 Liter

45 grams = X Liter

Cross-multiplying, we have:

[tex]47.46 \times X = 45 \times 1\\\\47.46X = 45\\\\X = \frac{45}{47.46}[/tex]

X = 0.95 grams

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