Calculate the amount of sodium carbonate (Na2CO3)(Na 2 CO 3 ) present in 200 ml of 2 M solution.
Question
Calculate the amount of sodium carbonate (Na2CO3)(Na 2 CO 3 ) present in 200 ml of 2 M solution.
Solution
Sure, here is the step by step solution:
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First, we need to understand the given information. We have a 2 M solution of sodium carbonate (Na2CO3) and the volume of the solution is 200 ml.
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Molarity (M) is defined as the number of moles of solute per liter of solution. So, a 2 M solution means there are 2 moles of sodium carbonate in 1 liter (or 1000 ml) of solution.
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To find the number of moles in 200 ml, we can set up a proportion. If 2 moles are in 1000 ml, then x moles are in 200 ml. So, 2/1000 = x/200.
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Solving for x gives us x = 2 * 200 / 1000 = 0.4 moles.
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Therefore, there are 0.4 moles of sodium carbonate in 200 ml of a 2 M solution.
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To find the amount of sodium carbonate in grams, we need to multiply the number of moles by the molar mass of sodium carbonate. The molar mass of Na2CO3 is approximately 105.99 g/mol.
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So, the amount of sodium carbonate is 0.4 moles * 105.99 g/mol = 42.396 grams.
Therefore, there are approximately 42.4 grams of sodium carbonate in 200 ml of a 2 M solution.
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