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4 years ago

What is the mole fraction of solute and the molal concentration for an aqueous solution that is 20.0% NaOH by mass?

1 answer:

8 0

Let's assume we have 100 grams of this solution, and therefore 20.0 grams of NaOH, and 80.0 grams of water.

NaOH has a molar mass of 40.00 grams/mole, so we can convert NaOH to moles: (20.0 grams of NaOH) * (1 mole NaOH/40.00 grams NaOH) = 0.500 moles NaOH.

Next, we have a molar mass of water of 18.02 g/mol, so we can convert water to moles:

80.0 grams H2O * (1 mole H2O/18.02 grams) = 4.44 moles H2O
The mole fraction of NaOH is the moles of NaOH over the total moles of all parts of the solution. Therefore:

(0.500 moles NaOH)/(0.500 moles + 4.44 moles) = 0.101
0.101 = mole fraction of solute

The molal concentration is defined as the number of moles of solute over the number of kilograms of solvent (water).
We have 0.500 moles of NaOH, and 0.0800 kg of water, so it becomes:
(0.500 moles NaOH)/(0.0800 kg H2O solvent) = 6.25 molal solution

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3 years ago

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3 years ago

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3 years ago

If 3.8 moles of zinc metal react with 6.5 moles of silver nitrate, how many moles of silver metal can be formed, and how many mo

bulgar [2K]

Answer: 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

Explanation:

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = $\frac{1}{2}\times 6.5=3.25moles$ of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = $\frac{2}{2}\times 6.5=6.5moles$ of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

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3 years ago