What mass in grams of MgSO4 is required to make 59.3 mL of 2.68 M
solution?

Answers

Answer 1

Answer:

Approximately [tex]19.1\; \rm g[/tex].

Explanation:

Number of moles of formula units of magnesium sulfate required to make the solution

The unit of concentration in this question is "[tex]\rm M[/tex]". That's equivalent to "[tex]\rm mol \cdot L^{-1}[/tex]" (moles per liter.) In other words:

[tex]c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}[/tex].

However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:

[tex]\displaystyle V = 59.3\; \rm mL = 59.3 \; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.0593\; \rm L[/tex].

Calculate the number of moles of [tex]\rm MgSO_4[/tex] formula units required to make this solution:

[tex]\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}[/tex].

Mass of magnesium sulfate in the solution

Look up the relative atomic mass data of [tex]\rm Mg[/tex], [tex]\rm S[/tex], and [tex]\rm O[/tex] on a modern periodic table:

[tex]\rm Mg[/tex]: [tex]24.305[/tex].[tex]\rm S[/tex]: [tex]\rm 32.06[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].

Calculate the formula mass of [tex]\rm MgSO_4[/tex] using these values:

[tex]M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}[/tex].

Using this formula mass, calculate the mass of that (approximately) [tex]0.159\; \rm mol[/tex] of [tex]\rm MgSO_4[/tex] formula units:

[tex]\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}[/tex].

Therefore, the mass of [tex]\rm MgSO_4[/tex] required to make this solution would be approximately [tex]19.1\; \rm g[/tex].


Related Questions

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Answers

Answer:

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Answers

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what mass of calcium chloride will be produced from 3.750 moles of hcl reacting with ca(oh)2

Answers

Answer:

208.125g

Explanation:

Given parameters:

 Number of moles of HCl = 3.75moles

Unknown:

 Mass of CaCl₂ that will be produced = ?

Solution:

The reaction equation is given as:

  2HCl  + Ca(OH)₂   →   CaCl₂ + 2H₂O

Now, to find the mass of CaCl₂, we have to work from the known to the unknown.

The known specie here is the number of moles of HCl;

        2 mole of HCl will produce 1 mole of  CaCl₂

       3.75moles of HCl will produce [tex]\frac{3.75}{2}[/tex]   = 1.875moles

The mass of  CaCl₂;

  Mass of  CaCl₂ = number of moles of  CaCl₂ x molar mass

Molar mass of  CaCl₂ = 40 + 2(35.5) = 111g/mol

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Answers

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Answers

Answer:

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Explanation:

The given values are:

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[tex]M_o=20 \ g/L\times V(L)[/tex]

     [tex]=20 \ V[/tex]

As we know,

⇒ [tex]Recovery \ percent=\frac{Amount \ of \ substance \ recovered}{Amount \ of \ substance \ originally \ taken}\times 100[/tex]

On substituting the given values, we get

                                [tex]=\frac{0.130}{20 \ V}\times 100[/tex]

                                [tex]=\frac{0.65}{V} \ percent[/tex]

Note: percent = %

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Answer:

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Explanation:

Answer:

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Answers

Answer:

It's B

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I got it right on edjenuity

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What color do you get if you mix red and blue light?
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Answers

Answer:

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I will comment more below.

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Answers

Answer:

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Answers

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Answers

Hey there !

Molecular Weight: 262.86 g/mol

Volume in liters : 1,000 mL => 1,000 / 1000 => 1 L

Number of moles of solution :

number of moles = volume x molarity

number of moles = 1 x 0.3  => 0.3 moles

Therefore:

1 mole Mg₃(PO₄)₂ ------------------- 262.86 g

0.3 moles ----------------------------- mass of Mg₃(PO₄)₂  ??

mass of Mg₃(PO₄)₂ = 0.3 x 262.86 / 1

mass of Mg(PO₄)₂ => 78.858 g

Hope this helps!

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