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3 years ago

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Identify the mole ratio of water to sodium metal? Determine the number of moles of Sodium that reacted and determine the number

of kg of Sodium that reacted. Sodium is 22.99 g/mol

1 answer:

frozen [14]3 years ago

7 0

Number of moles of sodium reacted is 2 moles. 0.04598 Kg is the mass of sodium that reacted.

Mole fraction is 1 mole ratio is 1:1

Explanation:

Balanced chemical reaction:

2 Na + 2H20 ⇒ 2NaOH + H2 (gas)

atomic mass of sodium = 22.99 g/mol

from the reaction it can be seen that 2 moles of sodium reacted to form 2 moles of sodium hydroxide soluition.

The mass of sodium reacted can be calculated from the formula:

mass = atomic mass of one mole x number of moles

mass = 22.99 x 2

= 45.98 grams is the mass of sodium, to convert it into kilograms it is divided by 1000

= 0.04598 Kg is the mass of sodium that reacted.

Mole fraction formula is:

$\frac{moles of solute}{moles of solution}$ = mole fraction ( solute Na = 2 moles, solution is NaOH =2 moles)

putting the values:

$\frac{2}{2}$

= 1 mole ratio is 1:1

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Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

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Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

i = van't Hoff factor

$k_b$ = Molal Elevation constant of solvent

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Molality of the solution = 0.19

Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

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i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

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Molality of the solution = 0.13

Elevation is boiling point solution :

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Molality of the solution = 0.31 m

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$\Delta T_b=1\times k_b\times 0.31 m$

$\Delta T_b=0.31 m\times k_b$

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The decreasing order of solution from highest boiling point to lowest boiling point is :

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