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dexteright02

Hello!

What the percent by mass of 5.0 g of iron (II) sulfate dissolved in 75.0 g of water?

We have the following data:

m1 (solute mass - iron II sulfate) = 5.0 g

m2 (solvent mass - water) = 75.0 g

m (solution mass) = m1 + m2 = 5.0 + 75.0 = 80.0 g

%m/m (percent mass by mass) = ?

We apply the data to the formula

\(\%\:m/m = \dfrac{m_1}{m}*100\)

\(\%\:m/m = \dfrac{5}{80}*100\)

\(\%\:m/m = 0.0625*100\)

\(\boxed{\boxed{\%\:m/m = 6.25}}\Longleftarrow(solute\:in\:percent)\:\:\:\:\:\:\bf\blue{\checkmark}\)

Answer:  

The percent by mass of solute and mass solution of Iron II Sulfate is 6.25%  

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CintiaSalazar

Considering the definition of percentage by mass, the percent by mass of 5.0 g of iron (II) sulfate dissolved in 75.0 grams of water is 6.25 %.

The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

\(percent by mass= \frac{mass of solute}{mass of solution}x100\)

In this case, you know:

  • mass of solute= 5 g
  • mass of water= 75 g
  • mass of solution= mass of solute + mass of water= 5 + 75 g= 80 g

Replacing:

\(percent by mass= \frac{5 g}{80 g}x100\)

Solving:

percent by mass= 6.25 %

Finally, the percent by mass of 5.0 g of iron (II) sulfate dissolved in 75.0 grams of water is 6.25 %.

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