Can you help me these 2 queations? if you can\'t see question 1 is: A stock solu
ID: 478035 • Letter: C
Question
Can you help me these 2 queations?
if you can't see question 1 is: A stock solution of NaCl is made using 25.738g NaCl and diluting to a final volume of 250.0 mL. what is the molar it of this stock solution?
Question 2: you use the stock solution above to make 3 dilute solutions. Each of the new solutions are diluted to a total of 150.0mL. What is the molarity of each new solution?
Amount of Stock NaCl solution used (In Chart)
25.0mL
15.0mL
1.0mL
beraton Questions Colerimetry is a technique used to determine the eoacentratae enakeelesExplanation / Answer
1) molarity of NaCl solution = ( mass of NaCl/ molar mass of NaCl) x (1/ volume in Litres)
= ( 25.738 g / 58.5 g/mol ) x (1/0.25 L)
= 1.76 M
Therefore,
molarity of NaCl stock solution = 1.76 M
2) Given that Each of the new solutions are diluted to a total of 150.0mL
a) 25 mL
M1 = 1.76 M , V1 = 25 mL , V2 = 150 mL , M2 = ?
M1V1 = M2V2
M2 = M1V1/ V2
= 1.76 M x 25 mL / 150 mL
= 0.293 M
Therefore, molarity of new solution = 0.293 M
b) 15 mL
M1 = 1.76 M , V1 = 15 mL , V2 = 150 mL , M2 = ?
M1V1 = M2V2
M2 = M1V1/ V2
= 1.76 M x 15 mL / 150 mL
= 0.176 M
Therefore, molarity of new solution = 0.176 M
c) 1.0 mL
M1 = 1.76 M , V1 = 1.0 mL , V2 = 150 mL , M2 = ?
M1V1 = M2V2
M2 = M1V1/ V2
= 1.76 M x 1.0 mL mL / 150 mL
= 0.0117 M
Therefore, molarity of new solution = 0.0117 M
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