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3 years ago

A rigid 3.80 L sealed vessel contains 0.650 mol Ne, 0.321 mol Kr, and 0.190 mol Xe. Find the density of the mixture in g/L.

Chemistry

1 answer:

Eddi Din [679]3 years ago

7 0

Answer:

17.09g/L

Explanation:

Density = total mass of elements/ volume

We need to find the mass of each mixture constituents using their molar mass:

mole = mass/molar mass

For Neon (Ne) which contains 0.650mol;

0.650 = mass/20.18

mass = 0.650 × 20.18

mass = 13.12g

For Krypton (Kr) which contains 0.321mol;

0.321 = mass/83.79

mass = 0.321 × 83.79

mass = 26.89g

For Xenon (Xe) which contains 0.190mol;

0.190 = mass/131.3

mass = 0.190 × 131.3

mass = 24.95g

Total mass = 13.12g + 26.89g + 24.95g = 64.96g

Density = total mass / volume

Density = 64.96g / 3.80L

Density of the mixture = 17.09g/L

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3 years ago

1) 0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution. 2) 734 grams of lithium sulfate are dissolved to m

OLEGan [10]

Answer:

1)Molarity of 0.5 Moles of Sodium Chloride in 0.05 Liters of solution is 10 M.

2)Molarity of 734 grams of lithium sulfate are dissolved to make 2500 mL of solution is 2.68 M.

3)Molarity of a solution is made by adding 83 grams of sodium hydroxide to 750 mL of water is 2.77 M.

4)Molarity of a solution that contains .500 mol HC₂H₃O₂ in 0.125 kg H₂O is 4 M.

5)Molarity of a solution that contains 63.0 g HNO₃ in 0.500 kg H₂O is 2 M.

6)The mass of water required to form 3.0 M solution by dissolving 0.5kg of C₂H₅OH is 3.62 kg.

Explanation:

Molarity is given as

$M=\dfrac{n}{V}$

Here

n is number of moles
V is the volume of the solution in liters

Using this all the values are calculated as follows:

1)Molarity of 0.5 Moles of Sodium Chloride in 0.05 Liters of solution is

$M=\dfrac{n}{V}\\M=\dfrac{0.5}{0.05}\\M=10$

So the molarity is 10 M.

2)Molarity of 734 grams of lithium sulfate are dissolved to make 2500 mL of solution is

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of lithium sulphate is as follows:

$MM of Li_2SO_4=2\times Li+S+4\times O\\MM of Li_2SO_4=2\times 7+32+4\times 16\\MM of Li_2SO_4=110$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{734}{110}\\n=6.67$

V is given as

$V_{L}=\dfrac{V_{mL}}{1000}\\V_{L}=\dfrac{2500}{1000}\\V_{L}=2.5\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{6.67}{2.5}\\M=2.68\ M$

So the molarity is 2.68 M.

3)Molarity of a solution is made by adding 83 grams of sodium hydroxide to 750 mL of water is

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of Sodium hydroxide is as follows:

$MM of NaOH=Na+O+H\\MM of NaOH=23+16+1\\MM of NaOH=40$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{83}{40}\\n=2.075$

V is given as

$V_{L}=\dfrac{V_{mL}}{1000}\\V_{L}=\dfrac{750}{1000}\\V_{L}=0.75\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{2.075}{0.75}\\M=2.77\ M$

So the molarity is 2.77 M.

4)Molarity of a solution that contains .500 mol HC₂H₃O₂ in 0.125 kg H₂O is

$M=\dfrac{n}{V}$

V is given as

$V_{L}=\dfrac{mass}{density}\\V_{L}=\dfrac{0.125}{1 kg/L}\\V_{L}=0.125\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{0.5}{0.125}\\M=4.0\ M$

So the molarity is 4.00 M.

5)Molarity of a solution that contains 63.0 g HNO₃ in 0.500 kg H₂O is

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of HNO₃ is as follows:

$MM of HNO_3=H+N+3\times O\\MM of HNO_3=1+14+3\times 16\\MM of HNO_3=63$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{63}{63}\\n=1.00$

V is given as

$V_{L}=\dfrac{mass}{density}\\V_{L}=\dfrac{0.5}{1 kg/L}\\V_{L}=0.5\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{1}{0.5}\\M=2.00\ M$

So the molarity is 2.00 M.

6)Mass of water must be used to dissolve 0.500 kg C₂H₅OH to prepare a 3.00 m solution is calculated as follows

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of C₂H₅OH is as follows:

$MM of C_2H_5OH=2\times C+6\times H+O\\MM of C_2H_5OH=2\times 12+6\times 1+16\\MM of C_2H_5OH=46$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{500}{46}\\n=10.87$

V required is given as

$M=\dfrac{n}{V}\\V=\dfrac{n}{M}\\V=\dfrac{10.87}{3}\\V=3.62\ L$

So the mass of water is given as

$mass=density\times Volume\\mass=1 kg/L \times 3.62 L\\mass=3.62\ kg$

So the mass of water required to form 3.0 M solution by dissolving 0.5kg of C₂H₅OH is 3.62 kg.

6 0

3 years ago