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Sladkaya [172]
3 years ago
15

Find the percent by mass of sodium chloride in a 1.20 M NaCl solution. The density of the solution is 1.06 g/mL

Chemistry
2 answers:
Gekata [30.6K]3 years ago
6 0

<u>Answer:</u> The mass percent of sodium chloride in solution is 6.616 %

<u>Explanation:</u>

Molarity is defined as the number of moles present in one liter of solution.

Mathematically,

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

We are given:

Molarity of NaCl solution = 1.20 M

This means that 1.20 moles of NaCl is present in 1 L or 1000 mL of solution

  • To calculate the mass for given number of moles, we use the equation:

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

Molar mass of NaCl = 58.44 g/mol

Moles of NaCl = 1.20 moles

Putting values in above equation, we get:

1.20mol=\frac{\text{Mass of NaCl}}{58.44g/mol}\\\\\text{Mass of NaCl}=(1.20mol\times 58.44g/mol)=70.13g

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.06 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.06g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.06g/mL\times 1000mL)=1060g

  • To calculate the mass percentage of sodium chloride in solution, we use the equation:

\text{Mass percent of sodium chloride}=\frac{\text{Mass of sodium chloride}}{\text{Mass of solution}}\times 100

Mass of solution = 1060 g

Mass of sodium chloride = 70.13 g

Putting values in above equation, we get:

\text{Mass percent of sodium chloride}=\frac{70.13g}{1060g}\times 100=6.616\%

Hence, the mass percent of sodium chloride in solution is 6.616 %

Kay [80]3 years ago
3 0

Answer:

6.61%

Explanation:

You know that the equation to find the mass percent is "mass(%) = (mass of solute)/(mass of solution)". Now, to find the mass of solute (NaCl), we know that it is given by 1.2 moles/liter and its molecular weight is 58.44 g/mol, so if you multiply those quantities the result will be 70.128 g/liter. And now we need to calculate the mass of solution, that is given by the density of solution multiplied by the volume (1000ml). Finally we just need to divide the mass of solute (70.128) by the mass of solution (1060) and multiply it by 100.

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Answer:

Empirical CHO

molecular C4H4O4

Explanation:

From the question, we know that it contains 41.39% C , 3.47% H and the rest oxygen. To get the % composition of the oxygen, we simply add the carbon and hydrogen together and subtract from 100%.

This means : O = 100 - 41.39 - 3.47 = 55.14%

Next is to divide the percentage compositions by their atomic masses.

C = 41.39/12 = 3.45

O = 55.14/16 = 3.45

H = 3.47/1 = 3.47

Now we divide by the smallest value which is 3.45. We can deduce that this will definitely give us an answer of 1 all through as the values are very similar.

Hence the empirical formula of Maleic acid is CHO

Now we go on to deduce the molecular formula.

To do this we need the molar mass. I.e the amount in grammes per one mole of the compound.

Now we can see that 0.378mole = 43.8g

Then 1 mole = xg

x = (43.8*1)/0.378 = 115.87 = apprx 116

[CHO]n = 116

(12 + 1 + 16]n = 116

29n = 116

n = 116/29 = 4

The molecular formula is thus C4H4O4

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Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) =&gt; CH 3OH(l) Using the following information: C(graphite) +
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Answer:

\Delta H for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

C(graphite)+O_{2}(g)\rightarrow CO_{2}(g) \Delta H_{1}= -393.5 kJ

2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l) \Delta H_{2}= (2\times -285.8)kJ

CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)  \Delta H_{3}= 726.4 kJ

---------------------------------------------------------------------------------------------------

C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)

\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ

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D

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I hope this is correct

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