Respuesta :
Answer: The mass of 97 % of NaOH solution required is 114.33 g
Explanation:
To calculate mass of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
We are given:
Density of 10 % solution = 1.109 g/mL
Volume of 10% solution = 1 L = 1000 mL (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
[tex]1.109g/mL=\frac{\text{Mass of }10\%\text{ solution}}{1000mL}\\\\\text{Mass of }10\%\text{ solution}=1109g[/tex]
The mass of 10 % solution is 1109 g.
To calculate the mass of concentrated solution, we use the equation:
[tex]c_1m_1=c_2m_2[/tex]
where,
[tex]c_1\text{ and }m_1[/tex] are the concentration and mass of concentrated solution.
[tex]c_2\text{ and }m_2[/tex] are the concentration and mass of diluted solution.
We are given:
[tex]c_1=97\%\\m_1=?g\\c_2=10\%\\m_2=1109g[/tex]
Putting values in above equation, we get:
[tex]97\times m_1=10\times 1109\\\\m_1=114.33g[/tex]
Hence, the mass of 97 % of NaOH solution required is 114.33 g
To prepare 1.00 L of 10.0% NaOH by mass solution, 114 g of solid NaOH (97.0% by mass) are required.
What is the percent by mass?
It is the mass of the component divided by the mass of the solution, multiplied by 100%.
- Step 1. Calculate the mass of 1.00 L of solution.
The density of the solution is 1.109 g/mL.
1.00 × 10³ mL × 1.109 g/mL = 1109 g
- Step 2. Calculate the mass of pure NaOH required.
The concentration of NaOH will be 10.0% by mass, that is, 10.0 g of NaOH per 100 g of solution.
1109 g Solution × 10.0 g NaOH/100 g Solution = 111 g NaOH
- Step 3. Calculate the mass of impure NaOH required.
The purity of NaOH is 97.0%.
111 g NaOH (pure) × 100 g NaOH (impure)/97.0 g NaOH (pure) = 114 g NaOH (impure)
To prepare 1.00 L of 10.0% NaOH by mass solution, 114 g of solid NaOH (97.0% by mass) are required.
Learn more about mass percent here: https://brainly.com/question/4336659
