University:
University of Arizona
Course:
CHEM 141 | General Chemistry Lecture I
Academic year:

2017
Views:

113

Pages:

1
Author:

CloudMaster

To solve a dilution question. So, how many milliliters of a three molar solution of sulfuric acid do I need to make 450 milliliters of a 0.100 molarity of sulfuric acid? Here, we can see that‚Ä need to dilute a concentrated solution of three molars to produce a 0.10 or 0.100 molarity. We are given the molarity of a more concentrated solution and the volume and molarity of a more dilute solution containing the same solute. So, we must calculate the volume of the concentrated solution needed to prepare the dilute solution. In other words, we need to calculate the initial volume. That we need is the concentrated solution to produce the derivative solution. So, to do this, we can use the equation M1V1 = M2V2, which stands for the initial molarity times the initial volume is equal to the final molarity times the final volume. This is the information that we have. Here, the problems are the final volume and the final concentration. We needed to make this solution, so the final volume will be 450 milliliters and the final concentration will be 0.10 molarity. While here, this is the initial concentration that will be mi. So, we need to rearrange this equation to obtain V1. So, Vi is going to be equal to MF DF divided by mi. Because here we're multiplying, we need to divide so we can cancel this in this representation and leave this initial volume by itself. So, here we have: Vi = (3 M)(0.10 M)/(450 mL). In this case, we don't have to look for the lead years, because once we solve this equation D, it is going to be milliliters. So we don't have to change first to build to two liters; you wouldn't need to do that. So here, we just do the rearrange and we cancel even submolar molarity, and we will have 15 milliliters. So this means that we need to add 15 milliliters. We need to add 15 milliliters of sulfuric acid to 435 milliliters of water. This will give us a total volume of 450 milliliters. The concentration of the solution will then be 0.10 M (molar) in sulfuric acid.