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xxMikexx [17]
2 years ago
8

a solution was made by mixing sodium chloride () and water (). given that the mole fraction of water is 0.980 in the solution ob

tained from 64.9 g , calculate the mass of sodium chloride used.
Chemistry
1 answer:
Mnenie [13.5K]2 years ago
4 0

The mass of sodium chloride used <u>was 1.17 grams</u><u>.</u>

The mole fraction can be calculated by way of dividing the number of moles of 1 factor of an answer via the full variety of moles of all the additives of an answer. it is mentioned that the sum of the mole fraction of all the components inside the solution has to be the same as one.

mass  of NaCl given =  64.9 g

mole  = mass/molar mass

          = 64.9 / 58.5

          =<u> 1.109</u>

a mole fraction of water =  0.980

mole fraction of NaCl = 1 - 0.980

                                   = <u>0.02</u>

1 mole of NaCl =  58.5

mass of NaCl  = 58.5 × 0.02

                       =<u> 1.17 gram</u>

Mole Fraction describes the range of molecules contained within one aspect divided through the whole range of molecules in a given combination. it's miles quite beneficial whilst two reactive-natured components are mixed collectively.

Learn more about mole fraction here:-brainly.com/question/29111190

#SPJ4

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The pH scale for acidity is defined by pH=−log10[H⁺] where [H⁺]is the concentration of hydrogen ions measured in moles per liter
masha68 [24]

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7 0
3 years ago
An intravenous infusion is to contain 15 mEq of potassium ion and 20 mEq of sodium ion in 500 mL of 5% dextrose injection. Using
Studentka2010 [4]

Answer:

To supply the required ions it is necessary to inject 5,6mL of 6g/30mL solution and 131,1 mL of 0,9% solution.

Explanation:

1mEq of sodium are 59mg of NaCl and 1mEq of potassium are 75mg KCl

in intravenous infusion 15 mEq of K are:

15x75mg KCl = 1,125g of KCl

And 20 mEq of Na are:

20x59mg NaCl = 1,18g of NaCl

To supply the potassium ion it is necessary to inject:

1,125g of KCl×\frac{30mL}{6g} =<em> 5,6mL  of 6g/30mL solution</em>

And, to supply the sodium ion it is necessary to inject:

1,18g of NaCl×\frac{100mL}{0,9g} = <em>131,1 mL of 0,9% solution</em>

<em />

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6 0
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