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maksim [4K]
3 years ago
7

The mole fraction of a non electrolyte (MM 101.1 g/mol) in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL.

Calculate the molarity of the solution.
Chemistry
1 answer:
yanalaym [24]3 years ago
8 0

Answer:

Molarity for the solution is 1.05 mol/L

Explanation:

Mole fraction of solute = 0.0194

Solution's density = 1.0627 g/mL

We must know that sum of mole fraction = 1

Mole fraction of solute + Mole fraction of solvent = 1

0.0194 + Mole fraction of solvent = 1

Mole fraction of solvent = 1 - 0.0194 → 0.9806

Molarity is mol of solute in 1L of solution, so we have to determine solution's volume in L

With molar mass we can determine the mass of solute and solvent and then, the solution's mass

0.0194 mol . 101.1 g/ mol = 1.96 g of non electrolyte solute

0.9806 mol . 18 g/mol = 17.65 g of water

Mass of solution = mass of solute + mass of solvent

1.96 g + 17.65 g = 19.6 g (mass of solution)

Solution's density = Solution's mass / Solution's volume

1.0627 g/mL = 19.6 g / Solution's volume

Solution's volume = 19.6 g / 1.0627 g/mol →18.4 mL

Let's convert the mass from mL to L

18.4mL . 1L / 1000 mL = 0.0184 L

We have the moles of solute, so let's determine molarity

mol/L → 0.0194 mol / 0.0184 L =  1.05 M

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A sucrose solution is prepared to a final concentration of 0.210 MM . Convert this value into terms of g/Lg/L, molality, and mas
vova2212 [387]

Answer:

1) 71.9 g/L

2) 0.221 m olal

3)  7.05% by mass

Explanation:

Step 1: Data given

Concentration of sucrose = 0.210 M

Molar weight of sucrose = 342.3 g/mol

Density of solution = 1.02 g/mL

Mass of water = 948.1 grams

Step 2: Convert this value into terms of g/L

(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L

Calculate the molality

Step 1: Calculate mass water

Suppose we have a volume of 1.00L

Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution

We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means

(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water

Step 2: Calculate molality

Molality = moles sucrose / mass water

(0.210 mol) / (0.9481 kg) = 0.221 mol/kg = 0.221 m olal

Mass %

% MAss = (mass solute / mass solution)*100%

(71.9 g) / (1020 g) *100% = 7.05% by mass

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