Molarity problems?

Question

I am having difficulty setting up these problems:

1. Calculate the number of grams of solute in 2.5L of a 3.00M HCl solution
2. Calculate the volume, in ml, of solution required to supply the following:2.00g of glucose from a 4.20M glucose solution

Yes, and I understand it involves moles/Liter

thanks for the leading, and I understand answers provided do not (sometimes) allow for learning. But, I am in review, not homework, for a test tomorrow, and these are practice questions. I spent 1 hour yesterday attempting to sort out this section: much longer than my allotted time for this preparation, and after attempting on my own to fight with my problems with dimensional analysis, I figured by being directed on a few answers, I could then practice the rest of the problems on my own. Thanks again for everyone's help with this....

Answer 1

Remember that Molarity = moles solute/liters solution.

(1) 3.00 M HCl = 3.00 mol HCl / 1 L soln
You have 2.5 L soln, so multiply 3.00 mol/L by 2.5 L and you will have 7.5 moles HCl due to the units canceling.

Now multiply the 7.5 mol HCl by the mass of HCl which is 36.5 g/mol. The units "mol" cancel and leave you with grams of HCl in the soln.

(2) Change the given mass of glucose to mol glucose by dividing the grams given by the molar mass of glucose (180 g/mol).

2.00 g / 180 g/mol = mol glucose.

Set up a proportion using the answer from the number of moles over "V" equals 4.20 M over 1 L (or 1000 mL).

[Remember that 4.20 M = 4.20 mol / 1L]

(answer in mol) / unknown V = 4.20 mol / 1000 mL

The units "mol" cancel and leave you with your answer in mL.

**Please Note**
I am a teacher and I like to lead you to the answer - not work out the problem for you and give you the answer. I feel people learn better by an explanation than they do from being given the answer.

Answer 2

1) Given :

Vol = 2.5 L
M = 3.0
Mwt of HCl = 36.45

Find : Grams of solute w/c is HCl


Grams of solute = Vol X M X Mwt of HCl

= (2.5 L ) X (3 mole /L ) ( 36.45 g/mol)

= 273.4 g

2)
Given :

wt of glucose = 2.00g
M = 4.20
Mwt of glucose = 180

Find : Volume

Vol = wt / ( M ) (Mwt )

Vol = 2.00g / (4.20moles/L) (180g/mole) =
=0.00265 L = 2.65 ml

Answer 3

Molarity = moles solute / V solution
1.
Moles solute = Molarity x V = 3.00 x 2.50 = 7.5 moles HCl
Molecular weight = 36.46 g/mol
7.5 mol x 36.46 g/mo = 273.4 g

2.
Glucose is C6H12O6
Molecular weight = 180 g/mol
2.00 / 180 = 0.0111 moles glucose
M = mol / V
V = mol / M = 0.0111 / 4.20 = 0.00264 L = 2.64 mL