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3 years ago

Assuming that seawater is a 3.50 mass % aqueous solution of NaCl, what is the molality of seawater?

Chemistry

1 answer:

Fantom [35]3 years ago

7 0

Hey there!

Consider 100 g of solution:

Mass of NaCl = 3.50% of mass of seawater

( 3.50 / 100 ) * 100 => 3.50 g

Number of moles as shown below:

Molar mass NaCl = 58.44 g/mol

n = Mass / molar mass

n = 3.50 / 58.44 => 0.059 moles of NaCl

Mass of sweater:

Mass of solution - Mass of NaCl

100 - 3.50 = 96.5 g

96.5 g in Kg :

96.5 / 1000 => 0.0965 Kg

Therefore ,calculate molality by using the following formula:

molality = number of moles of solute / mass of solution

molality = 0.059 / 0.0965

molality = 0.61 m

Hope That helps!

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Explanation:

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28) Consider a 21.0 mL sample of pure lemon juice with a citric acid (H3C6H5O7) concentration of 0.30M. a. How many moles of cir

Damm [24]

<h3>#a. Answer:</h3>

0.0063 mole

<h3>Solution and explanation:</h3>

We are given 21.0 mL citric acid with a concentration of 0.30 M

Part a requires we calculate the number of moles of citric acid.

We need to know how to calculate the concentration of a solution;

Concentration or molarity = Number of moles ÷ Volume of the solution

Thus;

Number of moles = Concentration × Volume

Hence;

Moles = 0.30 M × 0.021 L

= 0.0063 mole

<h3>#b. Answer</h3>

1.21 g citric acid

<h3>Solution</h3>

Part B

We are required to calculate the mass of citric acid in the sample

Number of moles of a compound is calculated by dividing its mass by its molar mass.

Molar mass of Citric acid = 192.124 g/mol

Moles of citric acid = 0.0063 mole

But; Mass = Number of moles × Molar mass

Mass of citric acid = 0.0063 mol × 192.124 g/mol

= 1.21 g citric acid

<h3>#c. Answer</h3>

4.167 mL

<h3>Solution:</h3>

Part C

We are required to determine the initial volume before dilution;

We have;

Initial concentration (M1) = 0.30 M

Final volume (V2) = 250 mL or 0.25 L

Final concentration (M2) = 0.0050 M

Using the dilution formula we can get the initial volume;

Therefore, since; M1V1 =M2V2

V1 = M2V2÷M1

= (0.0050 × 0.25)÷ 0.30

= 0.004167 L or

= 4.167 mL

Therefore, the initial volume of the solution is 4.167 mL

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3 years ago

Calculate the percentage of each element in acetic acid, hc2h3o2, and glucose, c6h12o6.

My name is Ann [436]

<em>Acetic acid, HC2H3O2</em>

First, calculate for the molar mass of acetic acid as shown below.
M = 1 + 2(12) + 3(1) + 2(16) = 60 g

Then, calculating for the percentages of each element.
<em> Hydrogen:</em>
P1 = ((4)(1)/60)(100%) = <em>6.67%</em>

<em> Carbon:</em>
P2 = ((2)(12)/60)(100%) = <em>40%</em>

<em>Oxygen</em>
P3 =((2)(16) / 60)(100%) = <em>53.33%</em>

<em>Glucose, C6H12O6</em>

The molar mass of glucose is as calculated below,
6(12) + 12(1) + 6(16) = 180

The percentages of the elements are as follow,
<em> Hydrogen:</em>
P1 = (12/180)(100%) = <em>6.67%</em>

<em>Carbon:</em>
P2 = ((6)(12) / 180)(100%) = <em>40%</em>

<em>Oxygen:</em>
P3 = ((6)(16) / 180)(100%) = <em>53.33%</em>

b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.

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