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sp2606 [1]
4 years ago
5

The particle size of solute particles does not affects the rate. Question 14 options: True False

Chemistry
1 answer:
stepan [7]4 years ago
8 0
The answer is false
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A student dissolves of 15 g aniline in of a solvent with a density of . The student notices that the volume of the solvent does
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The given question is incomplete. The complete question is ;

A student dissolves of 15 g aniline in 200 ml of a solvent with a density of 1.05 g/ml. The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.

Answer: The molarity is 0.81 M and molality is 0.82 m

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n= moles of solute  

V_s = volume of solution in ml = 200 ml

{\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{15g}{93g/mol}=0.16mol

Now put all the given values in the formula of molarity, we get

Molarity=\frac{0.16\times 1000}{200}=0.81M

Thus molarity is 0.81 M

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molarity=\frac{n\times 1000}{W_s}

where,

n = moles of solute

W_s = weight of solvent in g

{\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{15g}{93g/mol}=0.16mol

Mass of solution = Density\times Volume=1.05g/ml\times 200ml=210g

mass of solvent = mass of solution - mass of solute = (210 - 15) g = 195 g

Now put all the given values in the formula of molality, we get

Molality=\frac{0.16\times 1000}{195g}=0.82mole/kg

Therefore, the molality of solution is 0.82m

3 0
3 years ago
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