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3 years ago

If you needed to make 100 mL of a 0.2 M fruit drink solution from the 1.0 M fruit drink

Chemistry

1 answer:

telo118 [61]3 years ago

8 0

Answer:

We take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution.

Explanation:

Using the rule that: the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution. (MV) before dilution = (MV) after dilution. M before dilution = 1.0 M, V before dilution = ??? mL. M after dilution = 0.2 M, V after dilution = 100 mL. ∴ V before dilution = (MV) after dilution / M before dilution = (0.2 M)(100 mL) / (1.0 M) = 20.0 mL. So, we take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution

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Answer: The mass of potassium sulfate that can be produced is 73.88 grams

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To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For KOH:

Given mass of KOH = 23.8 g

Molar mass of KOH = 56.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of KOH}=\frac{23.8g}{56.1g/mol}=0.424mol$

The chemical equation for the reaction of KOH and potassium hydrogen sulfate follows:

$KHSO_4+KOH\rightarrow K_2SO_4+H_2O$

As, potassium hydrogen sulfate is present in excess. It is considered as an excess reagent.

KOH is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of KOH produces 1 mole of potassium sulfate

So, 0.424 moles of KOH will produce = $\frac{1}{1}\times 0.424=0.424moles$ of potassium sulfate

Now, calculating the mass of potassium sulfate from equation 1, we get:

Molar mass of potassium sulfate = 174.26 g/mol

Moles of potassium sulfate = 0.424 moles

Putting values in equation 1, we get:

$0.424mol=\frac{\text{Mass of potassium sulfate}}{174.26g/mol}\\\\\text{Mass of potassium sulfate}=(0.424mol\times 174.26g/mol)=73.88g$

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3 years ago

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nika2105 [10]

A. Using a whole solid in the reactants

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3 years ago

Read 2 more answers

A cylinder contains nitrogen and oxygen under 4.015 atm pressure. The partial pressure of N2 is 2.907 atm. What is the partial p

Aleksandr-060686 [28]

Answer: The partial pressure of the water vapor is 1.108 atm

Explanation:

According to Dalton's law, the total pressure of a mixture of gases is the sum of individual pressures exerted by the constituent gases.

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Given: $p_{total}$ =4.015 atm

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$p_{H_2O}$ = ?

Thus $4.015=2.907+p_{H_2O}$

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Thus the partial pressure of the water vapor is 1.108 atm

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3 years ago