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kirill [66]
3 years ago
10

Resuelve esto:

Chemistry
1 answer:
Misha Larkins [42]3 years ago
3 0

Answer:

37.

20

Explanation:

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A student dissolves 15.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.13/gmL. The student notices that the vo
Kruka [31]

<u>Answer:</u> The molarity and molality of sucrose solution is 0.146 M and 0.129 m respectively

<u>Explanation:</u>

  • <u>Calculating the molarity of solution:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of sucrose = 15 g

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 300 L

Putting values in above equation, we get:

\text{Molarity of sucrose solution}=\frac{15\times 1000}{342.3\times 300}\\\\\text{Molarity of sucrose solution}=0.146M

Hence, the molarity of sucrose solution is 0.146 M

  • <u>Calculating the molality of solution:</u>

To calculate the mass of solvent, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solvent = 1.13 g/mL

Volume of solvent = 300 mL

Putting values in above equation, we get:

1.13g/mL=\frac{\text{Mass of solvent}}{300mL}\\\\\text{Mass of solvent}=(1.13g/mL\times 300mL)=339g

To calculate the molality of solution, we use the equation:

\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

m_{solute} = Given mass of solute (sucrose) = 15 g

M_{solute} = Molar mass of solute (sucrose) = 342.3 g/mol

W_{solvent} = Mass of solvent = 339 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{15\times 1000}{342.3\times 339}\\\\\text{Molality of solution}=0.129m

Hence, the molality of sucrose solution is 0.129 m

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