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2 years ago

What is the molarity of a solution that contains 15.0g of naoh per 500.0 ml of solution

Chemistry

1 answer:

VARVARA [1.3K]2 years ago

8 0

Answer:

0.4 M

Explanation:

Molarity is defined as moles of solute, which in your case is sodium hydroxide,

NaOH

, divided by liters of solution.

molarity

moles of solute

liters of solution

Notice that the problem provides you with the volume of the solution, but that the volume is expressed in milliliters,

Moreover, you don't have the number of moles of sodium hydroxide, you just have the mass in grams. So, your strategy here will be to

determine how many moles of sodium hydroxide you have in that many grams

convert the volume of the solution from milliliters to liters

So, to get the number of moles of solute, use sodium hydroxide's molar mass, which tells you what the mass of one mole of sodium hydroxide is.

⋅

1 mole NaOH

40.0

0.175 moles NaOH

The volume of the solution in liters will be

500

⋅

1 L

1000

0.5 L

Therefore, the molarity of the solution will be

0.175 moles

0.5 L

0.35 M

Rounded to one sig fig, the answer will be

0.4 M

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1 year ago

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Based on the information provided, it appears that you will need to calculate the amount of heat absorbed by the water from the peanut that was burned. We are given the following information:

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3 years ago

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

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After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$

Thereby, the magnitude and direction of work turn out:

$W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ$

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3 years ago

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Artemon [7]

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What is the molarity of a solution that contains 15.0g of naoh per 500.0 ml of solution ​

What is the molarity of a solution that contains 15.0g of naoh per 500.0 ml of solution