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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
[tex]\%\:m/m = \dfrac{m_1}{m}*100[/tex]
[tex]\%\:m/m = \dfrac{5}{80}*100[/tex]
[tex]\%\:m/m = 0.0625*100[/tex]
[tex]\boxed{\boxed{\%\:m/m = 6.25}}\Longleftarrow(solute\:in\:percent)\:\:\:\:\:\:\bf\blue{\checkmark}[/tex]
Answer:
The percent by mass of solute and mass solution of Iron II Sulfate is 6.25%
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[tex]\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}[/tex]
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:
[tex]percent by mass= \frac{mass of solute}{mass of solution}x100[/tex]
In this case, you know:
Replacing:
[tex]percent by mass= \frac{5 g}{80 g}x100[/tex]
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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