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Offline blubbermcblubface

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Need some help on these questions
« on: April 08, 2020, 10:40:34 AM »
Hello all,

Sorry if these are super basic questions, we just started learning about solutions and concentrations in my class, but I was absent for most of this semester, and I am really, really bad at chemistry.

So here are my questions:
(1) Does 50ml of a 1M solution of Calcium Chloride contain twice as many Chloride Ions than 50ml of a 1M solution of Sodium Chloride?

(2) I need to use a solution of 3M HCl to make 100ml of 1.25M HCL. How many ml of 3m HCL should I use?

(3) If 35g of Magnesium Nitrate is dissolved to make 250ml of solution, what i the molarity of the final solution?

Thanks in advance!

Offline blubbermcblubface

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Re: Need some help on these questions
« Reply #1 on: April 08, 2020, 10:49:15 AM »
Sorry, I'm not sure if I need to show my work or not, so I will just in case.

For #1, I have absolutely no idea how to do the question, I don't know if there is a formula that I have to use, or which formula to use for it, and I am having a bit of trouble starting off the question.

For #2, I used the basic C1V1 = C2V2 and did (3M HCl)(x ml) = (1.25M)(100ml) and found that x = 41.7, I'm not too sure if that is correct.

For #3 I used the formula that states that molarity = (mol of solute)/(Kg of solvent). Here was my work flow for this question

    - Converting grams Mg(NO3)2 to mols Mg(NO3)2, and I got that 35g of Mg(NO3)2 is equivalent to 0.24 mols of Mg(NO3)2

    - From here I am stuck, because I do not know how to find the solvent, given that there is 250 ml of solution.

Offline sjb

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Re: Need some help on these questions
« Reply #2 on: April 08, 2020, 11:23:08 AM »
For #1, I have absolutely no idea how to do the question, I don't know if there is a formula that I have to use, or which formula to use for it, and I am having a bit of trouble starting off the question.

What happens when you dissolve CaCl2 and NaCl?


For #2, I used the basic C1V1 = C2V2 and did (3M HCl)(x ml) = (1.25M)(100ml) and found that x = 41.7, I'm not too sure if that is correct.

Do you mean 3m, or 3M (there is a difference)?

For #3 I used the formula that states that molarity = (mol of solute)/(Kg of solvent). Here was my work flow for this question

    - Converting grams Mg(NO3)2 to mols Mg(NO3)2, and I got that 35g of Mg(NO3)2 is equivalent to 0.24 mols of Mg(NO3)2

    - From here I am stuck, because I do not know how to find the solvent, given that there is 250 ml of solution.

Check your definition of molarity.

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