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Over [174]
3 years ago
8

3. What is the mass of HI used to create 2.50L of a 0.48 M solution of hydroiodic acid?

Chemistry
1 answer:
Daniel [21]3 years ago
5 0

Answer:

Mass = 153.48 g

Explanation:

Given data:

Volume of solution = 2.50 L

Molarity = 0.48 M

Mass required = ?

Solution:

Molarity = number of moles / volume in litter

Number of moles = Molarity × volume in litter

Number of moles = 0.48 M  × 2.50 L

Number of moles =  1.2 mol

Mass of HI:

Number of moles = mass/molar mass

Mass = Number of moles × molar mass

Mass = 1.2 mol × 127.9 g/mol

Mass = 153.48 g

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

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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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