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Karo-lina-s [1.5K]
2 years ago
12

How many grams of glucose (C6H12O6) are contained in 555 mL of a 1.77 M glucose solution?

Chemistry
1 answer:
Alex73 [517]2 years ago
8 0

176.980176 g of glucose (C_6H_{12}O_6) are contained in 555 mL of a 1.77 M glucose solution.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Molality = \frac{Moles \;solute}{Volume \;of \;solution \;in \;litre}

1.77 M = \frac{Moles \;solute}{0.555 L}

mole solute = 0.98235 g/mol

Moles  = \frac{mass}{molar \;mass}

0.98235 g/mol  = \frac{mass}{180.16 g/mol}

mass = 176.980176 g

Hence, 176.980176 g of glucose (C_6H_{12}O_6) are contained in 555 mL of a 1.77 M glucose solution.

Learn more about moles here:

brainly.com/question/8455949

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<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol

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Molarity of TRIS base solution = 0.2 M

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Putting values in above equation, we get:

0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol

Volume of solution = 50 + 60 = 110 mL = 0.11 L    (Conversion factor:  1 L = 1000 mL)

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pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})

We are given:

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[\text{TRIS base}]=\frac{0.012}{0.11}

pH = ?

Putting values in above equation, we get:

pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7

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