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Question: 40g MgO dissolves in \[{{H}_{2}}O~\] to form 200 ml of solution, if the density of solution is 1.5g/...

40g MgO dissolves in H2O {{H}_{2}}O~ to form 200 ml of solution, if the density of solution is 1.5g/ml. Calculate the molarity of the solution.

Explanation

Solution

We need to know the volume of the solution in order to know the molarity of the solution. By basic physics, we can relate density, volume and mass as density is defined as mass per unit volume of a substance. This relation is helpful in this question.

Complete step by step solution:
-Molarity is defined as the ratio of the moles of the solute to the volume of the solution and can be expressed mathematically as
molarity=moles of solutevolume of solution=molLmolarity=\dfrac{moles\text{ }of\text{ }solute}{volume\text{ of solution}}=\dfrac{mol}{L}
-In the given question, we are given the density of the solution and the weight of the solute. Also we are given the volume of the solution. We know that density, mass and the mass as all these three terms are related to each other as density=massvolumedensity=\dfrac{mass}{volume}
-But to find the molarity, we need the volume of the solution itself and not of the solvent which is given to us already. So we do not need the density. We just need to find the moles of the solute with the help of mass of solute.
-Moles of solute = wt. in gramsmolecular wt.=4024+16=4040=1 mole\dfrac{wt.\text{ in grams}}{molecular\text{ wt}\text{.}}=\dfrac{40}{24+16}=\dfrac{40}{40}=1\text{ }mole
So, molarity=moles of solutevolume of solutionmolarity=\dfrac{moles\text{ }of\text{ }solute}{volume\text{ of solution}}
= 1200×103\dfrac{1}{200 \times {{10}^{-3}}}
= 5M

Therefore the molarity of the solution is 5M.

Note: Always take into account the unit of all the quantities present in the question. Remember that 1M= mol/L. No other unit defines 1M. Also, keep into account the relationship between the units of mass and the units of volume. 1L = 1000ml and 1ml = 1cc. Also, volume has to be taken of the solution and the moles are taken of the solute.