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3 years ago

A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water

Chemistry

1 answer:

S_A_V [24]3 years ago

8 0

Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

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The activation energy for a(n) _____ is quite large and usually takes extra energy from the environment, it is normally not a na

Ira Lisetskai [31]

Answer:

The activation energy for an endothermic reaction is quite large and usually takes extra energy from the environment, it is normally not a natural spontaneous process.

Explanation:

Endothermic reactions require absorbing energy of the surrounding mainly in the form of heat.

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3 0

2 years ago

Read 2 more answers

A student sets up the following equation to convert a measurement.

Sholpan [36]

Answer:

–0.13 Pa.m²

Explanation:

From the question given above, the following data were obtained:

Measurement (Pa.mm²) = –1.3×10⁵ Pa.mm²

Measurement (Pa.m²) =?

We can convert from Pa.mm² to Pa.m² by doing the following:

1 Pa.mm² = 1×10¯⁶ Pa.m²

Therefore,

–1.3×10⁵ Pa.mm² = –1.3×10⁵ Pa.mm² × 1×10¯⁶ Pa.m² / 1 Pa.mm²

–1.3×10⁵ Pa.mm² = –0.13 Pa.m²

Thus, –1.3×10⁵ Pa.mm² is equivalent to –0.13 Pa.m².

6 0

3 years ago

PlEASE HELP! 40!

podryga [215]

Answer:

moles Na = 0.1114 g / 22.9898 g/mol=0.004846

moles Tc = 0.4562g /98.9063 g/mol=0.004612

mass O = 0.8961 - ( 0.1114 + 0.4562)=03285 g

moles O = 0.3285 g/ 15.999 g/mol=0.02053

divide by the smallest

0.02053/ 0.004612 =4.45 => O

0.004846/ 0.004612 = 1.0 => Tc

to get whole numbers multiply by 2

Na2Tc2O 9

Explanation:

Hope it right hope it helps

5 0

2 years ago

A single hydrogen atom has a mass of 1.67 × 10−24 g. A sodium atom has an atomic mass of 23. How many sodium atoms are required

statuscvo [17]

<h3>Answer:</h3>

2.55 × 10²² Na Atoms

<h3>Solution:</h3>

Data Given:

M.Mass of Na = 23 g.mol⁻¹

Mass of Na = 973 mg = 0.973 g

# of Na Atoms = ??

Step 1: Calculate Moles of Na as:

Moles = Mass ÷ M.Mass

Moles = 0.973 g ÷ 23 g.mol⁻¹

Moles = 0.0423 mol

Step 2: Calculate No, of Na Atoms as:

As 1 mole of sodium atoms counts 6.022 × 10²³ and equals exactly to the mass of 23 g. So, we can write,

Moles = No. of Na Atoms ÷ 6.022 × 10²³ Na Atoms.mol⁻¹

Solving for No. of Na Atoms,

No. of Na Atoms = Moles × 6.022 × 10²³ Na Atoms.mol⁻¹

No. of Na Atoms = 0.0423 mol × 6.022 × 10²³ Na Atoms.mol⁻¹

No. of Na Atoms = 2.55 × 10²² Na Atoms

<h3>Conclusion: </h3>

2.55 × 10²² sodium atoms are required to reach a total mass of 973 mg in a substance of pure sodium.

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3 years ago

What volume(in mL) of a 0.500 M fructose(C6H12O6) solution can be made from 11.3 g of fructose?

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2 years ago