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1 year ago

How many moles of solute would be dissolved in .500 kg of solvent to make a 2.50 molal NaOH solution?

1 answer:

3 0

Molality is a measure of concentration that relates the moles of solute to the kilograms of solvent, it is described by the following equation:

$Molality=\frac{molSolute}{kgSolvent}$

We are given the molality(2.50m) and kilograms of solvent(0.500kg), so we solve for moles of solute from the equation:

$molSolute=Molality\times kgSolvent$ $\begin{gathered} molSolute=2.50m\times0.500kg \\ molSolute=1.25molSolute \end{gathered}$

To make a 2.50molal NaOH solution would be needed 1.25moles of solute

Answer: 1.25 moles

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Find the normality of 0.321 g sodium carbonate in a 250 mL solution.

Radda [10]

Answer:

the normality of the given solution is 0.0755 N

Explanation:

The computation of the normality of the given solution is shown below:

Here we have to realize the two sodiums ions per carbonate ion i.e.

N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)

= 0.1886eq ÷ 0.2500L

= 0.0755 N

Hence, the normality of the given solution is 0.0755 N

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2 years ago

A water treatment plant applies chlorine for disinfection so that 10 mg/L chlorine is achieved immediately after mixing. The vol

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Answer: The mass of chlorine needed by the plant per day is $9.4625\times 10^8mg$

Explanation:

We are given:

Volume o water treated per day = 25,000,000 gallons

Converting this volume from gallons to liters, we use the conversion factor:

1 gallon = 3.785 L

So, $\frac{3.785L}{1\text{ gallon}}\times 25,000,000\text{ gallons}=9.4625\times 10^7L$

Amount of chlorine applied for disinfection = 10 mg/L

Applying unitary method:

For 1 L of water, the amount of chlorine applied is 10 mg

So, for $9.4625\times 10^7L$ of water, the amount of chlorine applied will be $\frac{10mg}{1L\times 9.4625\times 10^7L}=9.4625\times 10^8mg$

Hence, the mass of chlorine needed by the plant per day is $9.4625\times 10^8mg$

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2 years ago

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Correction *

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