Question

1) A student added 56.4 grams of barium nitrate to 894 grams of water. The resulting solution had a density of 1.12 g/ml. What is the weight perfect, molarity, and molarity of the solution?

2) A student evaporated 545 ml of a .523M sodium chloride solution down to a volume of 275 ml. What is the molarity of this final solution?

Answer 1

ⓛwenghr percent - heighr0t Solute X 100/ weitht or Solution weight ot coluhon (weight at solulit, waht solvent) 56-4一ㄧㄨ· 100 (564+894 -5.934%. , weighh percent Moland | :_ Mola rily is the concentration ota soluton expeses as tha mumber ot moles or Solure per line 04. solupǐon, 1000mL1 Libe Densih 12ml ", 1000 mL 1.12 X1000 solure 1忄 1120 into mole Nouw, we have to canvext sram Holax mass o G)26.31 Atomic, wh.9e lba= 131.33 Aromie ust. oy 1.0067 261.34 asoluhe makea 阜mole t120solutev 1120 mste = 4.235 mole . 1000 ml ve ltitre cantatna A235 mole .molarviry -429s 285 mole,.molanity

Moloi. :-. Molality is defined as the no, or mole。 ot a solute dissolved in 1 kg ot the solve o the soive nk Tt can be exprened an w x 100O 56401 Molad Molality-Max1000 564ー000-6 W12 내 gt solventーー8949 Molality 안 the so lutiom- 0.241 rtolal

2) We know -

V₁S₁ = V₂S₂  

V₁= volume of solution initially = 545 mL

V₂= volume of solution finally =275mL ( as the initial volume has been evaporated to 275 mL )

S₁ = strength of the solution initially = 0.523 M

S₂= strength of the final solution (we have to determine this )

putting all values in the equation we get--

V₁S₁=V₂S₂

545× 0.523 = 275 ×S₂

Hence S₂= 1.036 Molar

Molarity of the final solution is 1.036 Molar