Calculate the motility of a solution containing 14.3 g of NaCl in 42.2 g of wate
ID: 534817 • Letter: C
Question
Calculate the motility of a solution containing 14.3 g of NaCl in 42.2 g of water. A) 2.45 times 10^-4 m B) 4.80 times 10^-4 m C) 2.45 times 10^-1 m D) 103 m E) 5.80 m The density of a 20.3 M CH_3OH (methanol) solution is 0.858 g/mL. What is the molality of this solution? H_2O is the solvent. A) 17.4 m B) 20.8 m C) 23.7 m D) 70.0 m E) 97.6 m What mass of Li_3PO_4 is needed to prepare 500. mL of a solution having a lithium ion concentration of 0.175 M A) 6.75 g B) 10.1 g C) 19.3 g D) 30.4 g E) 3.38 gExplanation / Answer
Ans 1) Given Mass of solute (NaCl) = 14.3g
Molar mass of solute = 58.5g
Mass of solvent (H2O)=42.2g
Number of moles of solute = given mass /molar mass
= 14.3/58.5
= 0.244
Molality = number of moles of solute x1000 / mass of solvent
= 0.244x1000/42.2
= 5.80m
Ans 2) Given density of solution = 0.858g/ml
Volume of solution = 1000ml
mass of solution =volume x density
=1000 x 0.858
= 858g
Again Number of moles of CH3OH = 20.3
Molar mass of CH3OH = 32g
Mass of CH3OH = Number of moles x molar mass
= 20.3x32
= 649.6
Now Mass of solution = Mass of CH3OH + Mass of Water
mass of water = Mass of solution - Mass of CH3OH
=858-649.6
= 208.4
Molality = Number of moles of CH3OH x 1000 / Mass of water
=20.3 x 1000 / 208.4
=97.6M
Ans 3) Given
Number of moles of Li+ in 1L solution = 0.175
For 500ml = 0.5L solution
Number of moles of Li+ = 0.5 x 0.175
=0.087
Now The reaction will be
Li3PO4 ->3Li+ + PO43-
Number of moles of Li3PO4 = 0.087 / 3
=0.029
Molar mass of Li3PO4 = 115.79g
Number of moles = Mass of Li3PO4 / Molar mass
Mass of Li3PO4 = Number of moles x molar mass
= 0.029 x 115.79
=3.38g
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.