You are preparing some solutions for your instructor and have been asked to prepare 3.00 L of a 0.250M sodium hydroxide solution. To make the solution, you should weigh out: A. 30.0 g NaOH and add 3.00 L of water to it. B. 75.0 g NaOH and add water until the final solution has a volume of 3.00 L. C. 30.0 g NaOH and add water until the final solution has a volume of 3.00 L. D. 75 g NaOH and add 3.00 L water to it.

Question

Chemistry

Ashes

20 November, 23:38

You are preparing some solutions for your instructor and have been asked to prepare 3.00 L of a 0.250M sodium hydroxide solution. To make the solution, you should weigh out: A. 30.0 g NaOH and add 3.00 L of water to it. B. 75.0 g NaOH and add water until the final solution has a volume of 3.00 L. C. 30.0 g NaOH and add water until the final solution has a volume of 3.00 L. D. 75 g NaOH and add 3.00 L water to it.

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Answers (1)

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Melina Summers · Answer 1

C. 30.0 g NaOH and add water until the final solution has a volume of 3.00 L.

Explanation:

Molarity of a substance, is the number of moles present in a liter of solution.

M = n / V

M = molarity

V = volume of solution in liter,

n = moles of solute,

from the question,

M = 0.250M

V = 3.00 L

M = n / V

n = M * v

n = 0.250M * 3.00 L = 0.75 mol

Moles is denoted by given mass divided by the molecular mass,

Hence,

n = w / m

n = moles,

w = given mass,

m = molecular mass.

From the question,

n = 0.75 mol NaOH

m = molecular mass of NaOH = 40 g/mol

n = w / m

w = n * m

w = 0.75 mol * 40 g/mol = 30.0 g

Hence, by using 30.0 g of NaOH and dissolving it to make up the volume to 3 L, a solution of 0.250 M can be prepared.