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What is the molarity of an 85.0-mL ethanol (C2H5OH) solution

containing 1.77 g of ethanol?​

Respuesta :

Answer:

The molarity of the solution is 0,45 M.

Explanation:

We know that molarity is a concentration measure that expresses moles of solute per liter of solution (1000ml). First we proceed to calculate the mass of 1 mole of compound from the atomic weights of the elements of the periodic table and then we calculate the molarity by a simple rule of three:

Weight 1 mol C2H5OH= (Weight C)x2 + (Weight H)x6 + Weight 0

Weight 1 mol C2H5OH=12 g x 2 + 1g x 6 + 16 g= 46 g/mol

46 g ---1 mol ethanol

1,77g----x= (1,77g x  1 mol ethanol)/46 g= 0,038 mol ethanol

85 ml solution-----0,038 mol ethanol

1000ml solution---x= (1000ml solution x0,038 mol ethanol)/85 ml solution

x=0,45 M---> The solution is 0,45 M

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The molarity of an 85.0-mL ethanol (C2H5OH) solution containing 1.77 g of ethanol is 0.45M

MOLARITY:

  • The molarity of a solution can be calculated by dividing the number of moles by its volume. That is;

  • Molarity (M) = no. of moles (mol) ÷ volume (L)

  • According to this question, 85.0-mL ethanol (C2H5OH) solution containing 1.77 g of ethanol, 85.0-mL of an ethanol (C2H5OH) solution contains 1.77g of ethanol.

  • Molar mass of C2H5OH = 46g/mol

  • no. of moles = 1.77g ÷ 46g/mol

  • no. of moles = 0.0385mol

  • Volume of ethanol in L = 85/1000 = 0.085

  • Molarity of ethanol = 0.0385 ÷ 0.085

  • Molarity of ethanol = 0.45M

  • Therefore, the molarity of an 85.0-mL ethanol (C2H5OH) solution containing 1.77 g of ethanol is 0.45M.

Learn more about molarity: https://brainly.com/question/13386686?referrer=searchResults

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