whether the following solutions are saturated or unsaturated [show workingl 40.3
ID: 556475 • Letter: W
Question
whether the following solutions are saturated or unsaturated [show workingl 40.35 g isolved in 25.0 g h.o (Solubility of Caci-161.4 g100 g H,O) b 59 g NaCI dissolved in 25,0 g HO (Solubility of Nac-35.7 g/100gHo) many grams of NaOH (solute) must be added to 120 g of H2O to prepare a 20 % (n n) NaOH solution? Ho How many mL ofwater (solvent) is should be combined with 15.0gNaci to prepare a 10%(whv) NaCl solution 3. 4. How many grams of Na CO, are required to prepare 500 mL of a 1.25 M NaCO, solution? 5. Calculate the osmolarity of the following solutions: a. 0.1M KNO b. 0.35 M NaPO What volume of 5M NaOH solution is required to prepare 150 mL of 0.5M NaOH solution? What volume o water must be used to make up the solution? (2pts) 6. volume of sokute 100 "of sumo Volume Solution (m.L.) % (v/v) . mass souude +mass solventI velme solute + volume solvent _ %(m/v)= w (M)moles of solute Molarity (M-Liners of SolutionExplanation / Answer
a)Solubility = 40.35 gm/25 gm = 40.35*4/(25*4)= 161.4 gm/100 gm water
This is the solubility limit at the given temperature. Hence saturated.
b) 5.9 gm of NaCl/25 gm = 5.9*4/(25*4)= 23.6 gm/ 100 gm
but the actual solubility is 35.7 gm /100 gm of water. So the solution is unsaturated.
2. let x= mass of NaOH
Total mass of solution = mass of NaOH+ mass of water= x+120
Mass % of NaOH= 100*x/(x+120)= 20
Hence x/(x+120)= 0.2
, x= 0.2x+120*0.2, x=0.2x+24
0.8x= 24, x= 240/8= 30 gm
3. Assuming the density of 10% solution = 1 gm/cc ( since the solution is dilite)
Let x= mass of water
Mass of solution = x+15
Volume of solution = (x+15)*1= (x+15) ml
Volume % of NaCl = 100* volume of NaCl/ volume of solution = 10
100* 15/(x+15)= 10
15/(x+15)=0.1
15 =0.1x+1.5
13.5= 0.1x, x= 135 gm
4. moles of Na2CO3= molarity* volume in L= 1.25*500/1000 =0.625
mass of Na2CO3= moles * molar mass of Na2CO3= 0.625*106= 66.25 gm
5. An osmole (Osmol) is 1 mol of particles that contribute to the osmotic pressure of a solution.
for KNO3 dissociates into K+ ions and NO3- ions. So there are 2 osmoles in solution.
hence osmolarity = 2*0.1= 0.2
6. 0.5M Na3PO4 : Na3PO4 dissociates in to 3 Na+ ions and PO4-3 ion. So total ions in solution =4
osmolarity= 0.5*4= 2M
7. moles in 150 ml of 0.5M= molarity* volume in L= 0.5*150/1000 =0.075 moles
this many moles are there in 5M NaOH
Volume of NaOH in L=0.075/5= 0.015 L= 0.015*1000ml =15 ml of NaOH.
volume of water to be added= 150-15= 135 ml
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