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what is the mass%(m/m) of a solution prepared by dissolving 30.0 g of NaOH in 120.0 g of water​?

Sagot :

Percent by Mass

Discussion:

Percent by mass (also called percent by weight or weight percent) is the ratio of the mass of a solute to the mass of the solution, multiplied by 100. The formulas used in solving percent by mass are as follows:

[tex]\boxed{\text{percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} × 100}[/tex]

[tex]\boxed{\text{percent by mass} = \frac{\text{mass of solute}}{\text{mass of solute + mass of solvent}} × 100}[/tex]

The percent by mass is a unitless number because it is a ratio of two similar quantities.

[tex]\\[/tex]

Problem:

What is the mass%(m/m) of a solution prepared by dissolving 30.0 g of NaOH in 120.0 g of water?

Given:

[tex]\text{mass of solute = 30.0 g}[/tex]

[tex]\text{mass of solvent = 120.0 g}[/tex]

[tex]\text{solute} = \text{NaOH}[/tex]

[tex]\text{solvent} = \text{H}_{2}\text{O}[/tex]

Required:

[tex]\text{mass percent}[/tex]

Solution:

[tex]\text{mass percent} = \frac{\text{mass of solute}}{\text{mass of solute + mass of solvent}} × 100[/tex]

[tex]\text{mass percent} = \frac{\text{30.0 g}}{\text{30.0 g + 120.0 g}} × 100[/tex]

[tex]\boxed{\text{mass percent} = 20.0\%}[/tex]

[tex]\\[/tex]

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