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3 years ago

What is the molarity of a 10 L solution containing 5.0 moles of solute?

Chemistry

2 answers:

GalinKa [24]3 years ago

5 0

Explanation:

Number of moles consisting in a liter of solution is known as molarity.

Mathematically, Molarity = \frac{\text{no. of moles}}{\text{Volume in one liter}}[/tex]

It is given that volume is 10 liter and there is 5.0 moles of solute. Hence, we will calculate the molarity as follows.

Molarity = $\frac{\text{no. of moles}}{\text{Volume in one liter}}$

= $\frac{5.0 mol}{10 liter}$

= 0.5 mol/L

Thus, we can conclude that molarity of the given solution is 0.5 mol/L.

Setler79 [48]3 years ago

3 0

Molarity = Moles of solute/ L(liters) of solution

So let's plug in the information.

5.0 moles/10L = 0.5 M

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3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

How much heat is absorbed/released when 20.00 g of NH3(g) reacts in the presence of excess O2 (g) to produce NO (g) and H2O (l)

FrozenT [24]

Answer:

a. 342.9 kJ of heat are absorbed.

Explanation:

Calculation of the moles of $NH_3$ as:-

Mass = 20.00 g

Molar mass of $NH_3$ = 17.031 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{20.00\ g}{17.031\ g/mol}$

$Moles= 1.1743\ mol$

Given that:- $\Delta H=+1168\ kJ$

It means that 1 mole of $NH_3$ undergoes reaction and absorbs 1168\ kJ of heat

So,

1168 mole of $NH_3$ undergoes reaction and absorbs $1168\times 1168\ kJ$ of heat

Amount of heat absorbed = + 342.9 KJ

7 0

3 years ago