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2 years ago

What is the molality of a solution that has 30mg of K3PO4 dissolved in 40mL of water? (The density of water is 1.00 g/mL)

Chemistry

1 answer:

sweet [91]2 years ago

8 0

Answer:

m = 0.0035 m.

Explanation:

Hello there!

In this case, since the formula for the computation of the molality is:

$m=\frac{n_{solute}}{m_{solvent}}$

We can first compute the moles of solute, K3PO4 by using its molar mass:

$n=30mgK_3PO_4*\frac{1gK_3PO_4}{1000gK_3PO_4}*\frac{1molK_3PO_4}{212.27gK_3PO_4} =1.41x10^{-4}mol$

Next, since the volume of water is 40.0 mL and its density is 1.00 g/mL we infer we have the same grams (40.0 g). Thus, we obtain the following molality by making sure we use the mass of water in kilograms (0.04000kg):

$m=\frac{1.41x10^{-4}mol}{0.0400kg}\\\\m=0.0035m$

In molal units (m=mol/kg).

Best regards!

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Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What

Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $CH_4$ :-

Mass of $CH_4$ = 1.28 g

Molar mass of $CH_4$ = 16.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.28\ g}{16.04\ g/mol}$

$Moles_{CH_4}= 0.0798\ mol$

For $O_2$ :-

Mass of $O_2$ = 10.1 g

Molar mass of $O_2$ = 31.998 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{10.1\ g}{31.998\ g/mol}$

$Moles_{O_2}= 0.3156\ mol$

According to the given reaction:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

Limiting reagent is the one which is present in small amount. Thus, $CH_4$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0798\ moles= \frac{Mass}{44.01\ g/mol}$

Mass of $CO_2$ = 3.51 g

Theoretical yield = 3.51 g

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