Question

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.

Answer 1

Answer:

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 .