Answered

30) How many grams of CH3OH must be added to water to prepare 210 mL of a solution that is
1.50 M CH3OH?
A) 10.0g
B) 48.0 g
C) 3.20 g
D) 0.210 g
E) 22.08

Answer :

Neetoo

Answer:

Mass = 10.0 g

Explanation:

Given data:

Molarity = 1.50 M

Volume = 210 mL ( 0.21 L)

Grams of CH₃OH = ?

Solution:

Number of mole = molarity × volume

Number of mole = 1.50 M × 0.21 L

Number of mole = 0.315 mol

Mass of CH₃OH:

Mass = Number of moles ×  molar mass

Mass = 0.315 mol × 32.04 g/mol

Mass = 10.0 g

Cricetus

"10.0 g" CH₃OH must be added to water.

Chemical reaction:

According to the question,

Molarity = 1.50 M

Volume = 210 mL or, 0.21 L

Number of mole = Molarity × Volume

By substituting the values, we get

                            = [tex]1.50\times 0.21[/tex]

                            = [tex]0.315[/tex] mol

Mass of CH₃OH be:

Mass = Number of moles ×  Molar mass

         = [tex]0.315\times 32.04[/tex]

         = [tex]10.0[/tex] g

Thus the above solution is correct.

Find out more information about chemical reaction here:

https://brainly.com/question/24658038

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