Assistance please!

[tradanger]

I'm confused, can someone help me please?

A solution is prepared by mixing 25ml pentane (C₅H₁₂, density = 0.63g/cm³) with 45ml hexane (C₆H₁₄, density = 0.66g/cm³). Assuming that the volumes add on mixing, calculate:
a. the mass percentage
b. the molarity of the pentane

[Somye]

So you have two compounds which are combined

a) pentane has a density of .63g/mL and since you have 25ml, your mass is 15.75g
hexane has a density of .66h/mL and since you have 45mL, your mass is 29.7g
percentage comp is then putting one mass over the total mass, and multiplying by 100

b) your pentane mass is 15.75 grams, and the molar mass is 72g/mol
hence, you have .21875mol by m/M
c = n/v and since your volume is 70mL (25+45) you get a molarity of 3.125

[charmanderp]

The question was probably trying to trick you into thinking that there would be a reaction between pentane and hexane. As they're both completely saturated non-polar molecules they of course won't react and will just remain in solution.

[tradanger]

Thanks for the help guys! It seems so simple now ... Haha, I'll be back if I need anything more

[tradanger]

Hi guys, I need help with this question:
Calculate the mass of 10.00% stock solution required to prepare 0.8000kg of 4.000% solution.
I tried looking through my textbook and any chem notes I have but I can't find anything with percentages like this. Can someone please explain the question or the percentages?
Any help would be much appreciated

[Somye]

I'm assuming here your %s are w/v and that the solution is dissolved in water with density 1g/ml

therefore m(stock) = cv = .8 * 4/100 = 0.032g
so you need .032 of stock
therefore m(10% soln required) = 100/10 * 0.032 = .32L

So you'd need 320mL by my calculations..