A student dissolves 19.g of sucrose C12H22O11 in 425.mL of a solvent with a density of 0.82/gmL. The student notices that the volume of the solvent does not change when the sucrose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.

0.13M and 0.16m.

Explanation:

Molarity is defined as the moles of solute present in 1L of solution.

Molality is the moles of solute per kg of solvent.

To solve this problem we need to convert the mass of sucrose to moles using its molar mass and finding the volume in L of the solution and the mass in kg of solvent:

Moles sucrose:

Molar mass:

12C = 12*12.01g/mol = 144.12g/mol

22H = 22*1.005g/mol = 22.11g/mol

11O = 11*16g/mol = 176g/mol

Molar mass of sucrose is 144.12g/mol + 22.11g/mol + 176g/mol = 342.23g/mol

Moles are:

19.0g * (1mol / 342.23g/mol) = 0.0555 moles of sucrose

Liters solution:

425mL * (1L / 1000mL) = 0.425L

kg solvent:

425mL * (0.82g/mol) = 348.5g * (1kg / 1000g) = 0.3485kg

Molarity:

0.0555 moles / 0.425L

Molality:

0.0555 moles / 0.3485kg

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answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

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