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The concentration of glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},$ in normal spinal fluid is $\frac{75 \mathrm{mg}}{100 \mathrm{g}} .$ What is the molality of the solution?

$41.66 \times 10^{-3} \mathrm{m}$

Chemistry 102

Chapter 11

Solutions and Colloids

Solutions

Brown University

University of Toronto

Lectures

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Hello, everyone, this is Ricky and today we're working on Problem number 40 from Chapter 11 and so were given the amount of glucose and 100 grams of spinal fluid. And we have to calculate the morality. So first, to make our lives a little bit easier, let's do some unit conversions to remember that 75 milligrams of Samos saying 0.75 grams. And so if instead of dealing with just 100 grams of spinal fluid, um, we deal with 1000 grams, so we know supply, Billy, both top and bottom by 10 me get a 0.75 grams of glucose is present in one kilogram of spinal fluid. This just makes our calculation of it easier, because now essentially, the moles of glucose is equal to morality. Because you know, you divide anything by one. It's just it's it is itself. So let's calculate the moles of glucose by going 0.75 grams of C six h 12 06 Divide that by more mass which is equal to 180 and we get well 41 0.7 times 10 to the negative third moles, Then we can divide by our one kg and we'll get that. Our final answer is 41.7 times 10 to the negative three. Uh, morality for glucose books. Videos helpful, not see when the next.

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