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Nana76 [90]
3 years ago
10

Given that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have 100.mg of Na present? G

iven that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have of Na present? a. 0.256 b. 256 c. 39.0 d. 0.0390 e. none of the above
Chemistry
1 answer:
seraphim [82]3 years ago
8 0

<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.

<u>Explanation:</u>

We are given:

39.0 % of sodium in sodium chloride solution

This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution

Mass of sodium given = 100 mg = 0.1 g     (Conversion factor:  1 g = 1000 mg)

Applying unitary method:

If 39 grams of sodium metal is present in 100 grams of sodium chloride solution

So, if 0.1 grams of sodium metal will be present in = \frac{100}{39}\times 0.1=0.256g of sodium chloride solution.

Hence, the mass of sodium chloride solution present is 0.256 grams.

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The answer is <span>1.63 × 1024 atoms Fe.
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Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance:

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6.023 × 10²³ atoms : 1 mole = x : 2.70 moles
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3 years ago
A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of solution. She then took 15.0
yanalaym [24]

Answer: The concentration of KOH for the final solution is 0.275 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

 V_s = volume of solution in ml = 150 ml

moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles

Now put all the given values in the formula of molality, we get

Molality=\frac{0.178\times 1000}{150}=1.19M

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = molarity of stock solution = 1.19 M

V_1 = volume of stock solution = 15.0 ml

C_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 65.0 ml

Putting in the values we get:

1.19\times 15.0=M_2\times 65.0

M_1=0.275M

Therefore, the concentration of KOH for the final solution is 0.275 M

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