{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

122ch13c

# 122ch13c - 27 13.36 a Note the denominator in mass is the...

This preview shows pages 1–4. Sign up to view the full content.

27 13.36) a) Note the denominator in mass % is the total mass of solution Total mass = mass solute + mass solvent mass solute mass % = ------------------------ x 100% total mass solution 253.80 g I 2 ? mass I 2 = 0.035 mol I 2 x --------------- = 8. 8 83 g I 2 1 mol I 2 8. 8 83 g I 2 mass % I 2 = -------------------------------- x 100% (8. 8 83 g I 2 + 115 g CCl 4 ) = 7. 1 70 % = 7.2 % I 2 Note : Think of % as parts per hundred (it makes ppm, ppb and ppt easier to understand). b) mass solute ppm = ------------------------ x 10 6 total mass solution 0.0079 g Sr 2+ ppm Sr 2+ = -------------------------------- x 10 6 = 7. 8 99 = 7.9 ppm Sr 2+ 0.0079 g Sr 2+ + 10 3 g H 2 O Since the amount of Sr 2+ is so minute compared to the amount of H 2 O we could ignore it in the denominator in part b above and get the same result to 2 sig. fig. .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
28 13.37) A solution contains 14.6 g CH 3 OH in 184 g H 2 O. a) Find mol fraction of CH 3 OH, X CH3OH mol CH 3 OH X CH3OH = ----------------------- (No unit) total mol solution 1 mol CH 3 OH ? mol CH 3 OH = 14.6 g CH 3 OH x --------------------- = 0.45 5 6 mol CH 3 OH 32.04 g CH 3 OH 1 mol H 2 O ? mol H 2 O = 184 g H 2 O x ------------------ = 10. 1 2 mol H 2 O 18.016 g H 2 O 0.45 5 6 mol CH 3 OH X CH3OH = --------------------------------- = 0.042 7 1 = 0.0427 (0.45 5 6 + 10. 1 2) mol soln b) Find mass % CH 3 OH (also called wt % or wt/wt %) mass CH 3 OH mass % CH 3 OH = ------------------------ x 100% total mass solution 14.6 g CH 3 OH mass % CH 3 OH = ------------------------------------- x 100% (14.6 g CH 3 OH + 184 g H 2 O) = 7.3 5 14 % = 7.35 % CH 3 OH c) Find molality of CH 3 OH, m mol CH 3 OH m = ----------------- kg H 2 O 0.45 5 6 mol CH 3 OH m = -------------------------- = 2.4 7 65 = 2.48 m CH 3 OH 0.184 kg H 2 O
29 13.39) mol solute molarity, M = --------------- L solution a) Molarity of 0.540 g Mg(NO 3 ) 2 in 250.0 mL (0.2500 L) solution mol Mg(NO 3 ) 2 0.540 g Mg(NO 3 ) 2 1 mol Mg(NO 3 ) 2 10 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 11

122ch13c - 27 13.36 a Note the denominator in mass is the...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online