
If water is added to $ 50ml $ of a $ 0.04M $ solution so that it fills a $ 200ml $ beaker, what is the final concentration?
Answer
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Hint: Given that $ 50ml $ of a volume of a $ 0.04M $ solution is taken, from this the number of moles can be determined, and this volume is filled up to $ 200ml $ which is the final volume. By substituting the determined number of moles and final volume in the below formula gives the final concentration.
$ M = \dfrac{n}{V} $
$ M $ is final concentration
$ n $ is number of moles of final solution
$ V $ is the final volume of the solution.
Complete answer:
Molarity is also known as molar concentration. It is one of the units used to express concentration. It was represented by $ M $ . It has the units of $ mol{L^{ - 1}} $ .
The number of moles is the ratio of mass to molar mass in general, but in terms of molarity, it is the product of molarity and volume.
The number of moles of final solution will be the product of the given concentration and final volume will be
$ n = 50 \times {10^{ - 3}} \times 0.04moles $
The final volume is $ 200 \times {10^{ - 3}}L $
Substitute the number of moles, and final volume in the above formula, to obtain the final concentration
$ M = \dfrac{{50 \times {{10}^{ - 3}} \times 0.04moles}}{{200 \times {{10}^{ - 3}}L}} $
By further simplification, the value of final concentration will be $ 0.01mol{L^{ - 1}} $
Thus, water is added to $ 50ml $ of a $ 0.04M $ solution so that it fills a $ 200ml $ beaker, then the final concentration is $ 0.01mol{L^{ - 1}} $ .
Note:
While calculating the molarity or related terms from molarity, the volume of solution must be taken in litres only, if it is in millilitres, multiply the obtained molarity by $ 1000 $ or convert the millilitres into volume by taking $ 1litre $ equal to $ 1000ml $ .
$ M = \dfrac{n}{V} $
$ M $ is final concentration
$ n $ is number of moles of final solution
$ V $ is the final volume of the solution.
Complete answer:
Molarity is also known as molar concentration. It is one of the units used to express concentration. It was represented by $ M $ . It has the units of $ mol{L^{ - 1}} $ .
The number of moles is the ratio of mass to molar mass in general, but in terms of molarity, it is the product of molarity and volume.
The number of moles of final solution will be the product of the given concentration and final volume will be
$ n = 50 \times {10^{ - 3}} \times 0.04moles $
The final volume is $ 200 \times {10^{ - 3}}L $
Substitute the number of moles, and final volume in the above formula, to obtain the final concentration
$ M = \dfrac{{50 \times {{10}^{ - 3}} \times 0.04moles}}{{200 \times {{10}^{ - 3}}L}} $
By further simplification, the value of final concentration will be $ 0.01mol{L^{ - 1}} $
Thus, water is added to $ 50ml $ of a $ 0.04M $ solution so that it fills a $ 200ml $ beaker, then the final concentration is $ 0.01mol{L^{ - 1}} $ .
Note:
While calculating the molarity or related terms from molarity, the volume of solution must be taken in litres only, if it is in millilitres, multiply the obtained molarity by $ 1000 $ or convert the millilitres into volume by taking $ 1litre $ equal to $ 1000ml $ .
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