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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 13 - Problem 114p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 13 - Problem 114p

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# What is the molarity of an aqueous solution that lis 6.75% gucose (C6H12O6) by mass? ISBN: 9780321910295 34

## Solution for problem 114P Chapter 13

Introductory Chemistry | 5th Edition

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Problem 114P

Problem 114P

What is the molarity of an aqueous solution that lis 6.75% gucose (C6H12O6) by mass? (Assume a density of 1.03 g/ mL for the solution.)

Step-by-Step Solution:
Step 1 of 3

In this problem our goal is to know the molarity of a solution of glucose.

Molarity expresses the concentration of a solution as the number of mole in one liter of solution:

M=n/L

The only way to know how many moles of glucose are contained in 1 liter of solution is knowing first the mass of glucose contained in the same quantity of solution. Then we will divide the mass of glucose for its molecular weight to find the moles.

But how can we find the mass of glucose in one liter of solution.

The text gives us a concentration but it is by mass, so we must use the density of the solution to know the mass of one liter of solution.

Then you just need to follow the previous instructions to solve the problem.

How to solve it:

1) mass of 1 Liter of solution: density=Mass/volume, to find the mass

Mass=density⋅volume→1.03g/ml⋅1000ml=1030g of solution.

2) mass of glucose in 1 liter of solution (which means 1030 grams of solution): 6.53g:100=x:1030g

this is a proportion that means: 6.53 grams of glucose are contained in 100 grams of solution as x grams of glucose are contained in 1030 grams of solution (1 liter of solution)

x=1030⋅6,53/100=67,26g of glucose

3) how many moles are in 67.26 grams of glucose

MW (molar weight of glucose):180 g/ mol

moles=mass/molar weight=67.26/180=0.37 mol.

this are the moles of glucose contained in 1 liter of solution so according to the definition of molarity....the molarity is 0,37M.

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

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