Question About #5 on Post Module

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Rhiannon Imbeah 2I
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Joined: Fri Sep 25, 2015 3:00 am

Question About #5 on Post Module

Postby Rhiannon Imbeah 2I » Mon Jul 18, 2016 8:54 pm

Hello,

I am revisiting the Molarity and Dilution module. Question 5 in the post-module section is asking for the final molarity. After I finished converting the volume in mL to L, I was moving on to the molar calculations and the measurement is given in grams. Is the next step to convert the grams into the molar mass and then use that number to find the final molarity?

Thank you,

My work so far:
M(int)V(int)=M(fin)V(fin)
V(int)= 75mL-> 75mL/1000-> 0.075L
V(fin)= 125 mL -> 125mL/1000-> 0.125L
M(int)= 55.1 g of KCl

*Just to clarify: We convert the grams of KCl to moles?

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Re: Question About #5 on Post Module

Postby Chem_Mod » Tue Jul 19, 2016 10:55 am

Remember M is molarity (mole/L) and V is solution volume (L).

The answer is:

molar mass of KCl = 39.10 + 35.45 = 74.55 g/mol

# of moles = 55.1/74.55 = 7.39 x 10-1 mol

molarity = moles/vol = 7.39 x 10-1 mol/125 x 10-3 L = 5.91 mol/L


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