molality and mass problem

[inaccensa]

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What is the molality of a stock solution that is 10%
SDS by mass? SDS C11H23CH2OSO3–Na+ (MW-288) (STOCK SOLN-10%)
A. 0.028 m
B. 0.100 m
C. 0.347 m
D. 0.385 m

can someone plz explain how to do this
so 10% SDS is 10g in a 100gm sample or 10kg in 100kg sample
10/100* 288g =28.8gms
The remaining 90g is water
for molality
# moles /Kg solvent

(28.8 gms/288gms)*1 mole = 0.1mole

anyways my solution is way off 😕

 

[Bruinsfinalfour]

What is the molality of a stock solution that is 10%
SDS by mass? SDS C11H23CH2OSO3–Na+ (MW-288) (STOCK SOLN-10%)
A. 0.028 m
B. 0.100 m
C. 0.347 m
D. 0.385 m

can someone plz explain how to do this
so 10% SDS is 10g in a 100gm sample or 10kg in 100kg sample
10/100* 288g =28.8gms
The remaining 90g is water
for molality
# moles /Kg solvent

(28.8 gms/288gms)*1 mole = 0.1mole

anyways my solution is way off 😕

use 1 kg for the entire solution. 10% of 1 kg is 100 g. You have a 100 g of SDS. 100g /288(g/mole) = .347 moles SDS. Kg solvent = 1kg - 100 g = 900 g. Molality = moles SDS/ kg of solvent = .347 / .9 = 0.385. Answer should be D.

 

[inaccensa]

ur explanation is simpler than kaplan's👍