Tough Stoichiometry Question - Molarities, etc? (check my work)?

Question

Question: (a) volume in liters of 2.26 M potassium hydroxide that contains 8.42 g of solute. (b) number of Cu2+ ions in 52L of 2.3 M copper(II) chloride. (c) molarity of 275 mL of solution containing 135 mmol of glucose.

My Work:
(a) 8.42g KOH x (1mol/56g) = .150 mol KOH

2.26mol/L / .150 mol = 15.1 L --- answer

(b) 2.3 mol/1 L x 52L = 119.6 mol CuCl2
119.6 mol CuCl2 x (1mol Cu2+ ions/ 1 mol CuCl2) x (6.022e23 ions/ 1 mol Cu2+) = 7.20e25 ions -- answer

(c) 275 mL x (1L/1000 mL) = .275 L
135mmol x (10^-3mol/ 1mmol) = .135 mol

.135 mol/.275 L = .491 M glucose -- answer

if you see anything wrong PLEASE correct it! thanks

Answer 1

a) 1st part OK 2nd part NO!

0.150mole/(2.26mole/l) = 0.0663 l (66.3 ml sounds right for 8 g doesn't it!)

b) ok

c) yes

Well done!

Answer 2

I didn't check the calculation but be careful of the significant figures. The answers in (a) should contain 2 sig. fig. because you used 56 g/mol, if you have a better FW, then you can have 3 sig. fig. the most.
For (b), you should have only 2 sig. fig. becasue of 52L and 2.3M, they both contain 2 sig. fig., therefore, you can't have 3 sig. fig. in your answer.

Answer 3

Assuming that all and sundry the Pb(NO3)2 is fed on, all you're able to do is locate out how many moles of Pb(NO3)2 you have: .250M*.0100L= .0025 moles of Pb(NO3)2. on account that LiBr has a coefficient of two, you're able to multiply the moles of Pb(NO3)2 by way of 2: .0025 moles *2=.0.5 moles of LiBr. Divide moles of LiBr by way of liters of LiBr: .005moles/.0250L= .2M LiBr.

Answer 4

looks good to me