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The number of moles of sodium hydroxide present in 2.5L and 0.5M aqueous solution is?
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The number of moles of sodium hydroxide present in 2.5L and 0.5M aqueo...
Calculation of Moles of Sodium Hydroxide Present in Aqueous Solution


Given Information:


  • Volume of Solution (V) = 2.5 L

  • Concentration of Solution (C) = 0.5 M



Formula Used:


  • Moles (n) = Concentration (C) x Volume (V)



Calculation:


  • Substitute the given values into the formula

  • n = 0.5 M x 2.5 L

  • n = 1.25 moles



Answer:


  • The number of moles of sodium hydroxide present in 2.5L and 0.5M aqueous solution is 1.25 moles.



Explanation:

Moles of a substance can be calculated by multiplying its concentration (in moles per liter) and volume (in liters). In this problem, the given concentration is 0.5 M (moles per liter) and volume is 2.5 L. By multiplying these values, we get the number of moles of sodium hydroxide present in the solution, which is 1.25 moles.
Community Answer
The number of moles of sodium hydroxide present in 2.5L and 0.5M aqueo...
Molarity = no.of moles of solute/ vol of sol in L here molarity is 0.5 and vol = 2.5L...so i gues u can easily find out no. of moles of NaOH....
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Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced.This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200; 1 Faraday = 96,500 coulombs)If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is Correct answer is '446'. Can you explain this answer?

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The number of moles of sodium hydroxide present in 2.5L and 0.5M aqueous solution is?
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