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sashaice [31]
3 years ago
7

How many grams of solute are needed to make 37.5 mL of 0.750 M KI solution? Round to three significant digits.

Chemistry
1 answer:
Marysya12 [62]3 years ago
8 0

4.648 gm of solute is needed to make 37.5 mL of 0.750 M KI solution.

Solution:  

We will start with the Molarity  

\text { Molarity }=\text { Mole of solute } \div \text { liter of solution }

Also we know 1000 ml = 1 L

Therefore 37.5 ml by 1000ml we obtained 0.0375L  

Equation for solving mole of solute

\text { Mole of solute }=\text { Molarity } \times \text { Liters of solution }

Now, multiply 0.750M by 0.0375

Substitute the known values in the above equation we get

0.750 \times 0.0375=0.0281

Also we know that Molar mass of KI is 166 g/mol

So divide the molar mass value to get the no of grams.

0.028 \times 166=4.648

So 4.648 gm of Solute is required for make 37.5 mL of 0.750 M KI solution.

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Chlorine has two naturally stable isotopes: 35Cl (34.968853 amu) and 37Cl (36.965903 amu). The natural abundance of each isotope
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Answer:

35Cl ⇒ 34.968853 amu  has an abundance of 30.05%

Explanation:

The molar mass of chlorine, (which is the average of all its naturally stable isotope masses), is 36.36575 amu.

There are 2 naturally stable isotopes, this means together they have an abundance of 100%

The isotopes are:

35Cl ⇒ 34.968853 amu  has an abundance of X %

37Cl ⇒ 36.965903 amu  has an abundance of Y %

X + Y = 100%   OR X = 100% - Y

36.36575 = 34.968853X + 36.965903Y  

36.36575 = 34.968853(1-Y) + 36.936.96590365903Y

36.36575 = 34.968853 -34.968853Y + 36.965903Y

1.396897 = 1.99705Y

Y = 0.699 = 69.95%

X = 100-69.9 = 30.05%

To control, we can plug in the following equation:

34.968853 * 0.3005 + 36.965903 * 0.6995 = 36.3658

This means

37Cl ⇒ 36.965903 amu  has an abundance of 69.95 %

35Cl ⇒ 34.968853 amu  has an abundance of 30.05%

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