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Question

Question: The number of moles of solute present in \(600mL\) of \(0.05M\) solution is \(0.03mol\). A: True ...

The number of moles of solute present in 600mL600mL of 0.05M0.05M solution is 0.03mol0.03mol.
A: True
B: False

Explanation

Solution

Molarity is the measure of number of moles of a solute that are present in one liter of a solution. Number of moles of a substance can be calculated by dividing the given mass of a substance with the molecular mass.

Formula used: Molarity=number of moles ×1000volume of solution in mL = \dfrac{{{\text{number of moles }} \times 1000}}{{{\text{volume of solution in mL}}}}

Complete step by step answer:
In this question we have given the molarity of a substance and volume of its solution and we have to verify the number of moles of solute. Molarity is defined as the number of moles of a solute that are present in one liter of a solution. The units of molarity is MM. Number of moles of a substance can be calculated by dividing the given mass of a substance with the molecular mass of that substance. Formula to calculate molarity is,
Molarity=number of moles ×1000volume of solution in mL = \dfrac{{{\text{number of moles }} \times 1000}}{{{\text{volume of solution in mL}}}}
Molarity of solution is 0.05M0.05M (given)
Volume of solution is 600mL600mL (given)
Substituting these values in the above formula we can calculate the number of moles of solute. So,
0.05=number of moles ×10006000.05 = \dfrac{{{\text{number of moles }} \times 1000}}{{600}}
Solving this we get,
Number of moles=0.05×6001000=0.03 = \dfrac{{0.05 \times 600}}{{1000}} = 0.03
Therefore, the number of moles of solute are 0.030.03. Number of moles of solute that are given in the statement is 0.03mol0.03mol which is equal to the number of calculated moles. Therefore, the given statement is correct. So, the correct answer is option A that is true.

Note:
Molality is defined as the number of moles that are present per kilogram of the solvent. In the calculation of molality the mass of solvent is used and in the calculation of molarity the volume of solution is used.