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.

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<span>Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
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2 years ago

If liquid carbon disulfide reacts with 450 mL of oxygen to produce the gases carbon dioxide and sulfur dioxide, what volume of e

ivolga24 [154]

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<h3>Stoichiometric calculation</h3>

The reaction between liquid carbon disulfide and oxygen is represented by the equations below:

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The mole ratio of oxygen to carbon dioxide and sulfur dioxide produced is 3:1:2.

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Also for 450 mL of oxygen, 2/3 x 450 = 300 mL of sulfur dioxide will be required.

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<h3>Further explanation</h3>

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$\large{\boxed{\boxed{\bold{C_nH_{2n}O}}}$