Answer to Question #298434 in General Chemistry for Mitsu

Question

using conversion method, calculate the molality and molarity of aqueous phosphoric acid solution 1.88 gram/ml containing 20% by mass of h3 po4? h=1 p=31,o=16

Answer 1

1msample =msolute -msolvent

100-20=80

1kg=1000g so 1.88g will be 0.00188kg

RFM OF h3pO4=98

So the moles will be 20/98 =0.2040816327

Molality will be 0.204081/0.00188 =108.55

=108.55