Question

a) 2.00 (IV) It is desired to prepare 0.250 L of a 0.100 M solution of NaCl from a 1.00 M stock solution. How many mL of the stock solution must be taken for the dilution? (V) What volume of a 0.500 M HCl solution is needed to neutralize 10.0 mL of a 0.300 M NaOH solution. (VI) What volume of 0.416 M Mg(NO3)2 should be added to 255 mL of 0.102 M KNO, to produce a solution with a concentration of 0.278 M NO, ions? Assume volumes are additive. (VII) Calculate the osmotic pressure of 0.50 M sodium chloride at 25°C.

Answer 1

Answer page:01 (VV) stock solution must be taken for the dilution the tu coris dilution calculator is based on the following equation. Concentration (start) x volume (start) = Concentration (final) xvolume(final) This equation is commonly abboeviated as eive =C2V2 C2 = 0,100M V2 -0.250L Ci=1.00 M above.. Substituting these values on 1.00mxv= 0.1000x0.2500 = 0.025L 1.00x V V, = 0.025 -0.0251 N = 25 mL

page:02 - (V) Volume of 0.500 MHd solution is needed to neutralize 10.0 mL of 0.300M Naoh solution M, = 0. 300 M NAOH V = 10.0 mL000L M2= 0.500 M HCl V2 = ? Max, = M₂ V₂ ] V₂ = M, M2 Substitute given values on () we get = 0.300 st X0.0104 0.500 = 0.003 L 0.500 = 6X10². ₂ = 6 me 6 me Hcl oequired to neutralize 10 ml Naol solution

(VI) page.:03 concentration of Mg (NO3)2 = 0.416M concentration & KNO3 = 0.102 M volume of KNO3 = 955 ml 0.102 mok nog i number of moles of KNOX = 255 ml x K 1000 ml =.255 x X 0.102 mol KNO3 1000 = 0.0260 md knog The equation for the dissociation of knog is as follows KNO, (aq) -->** (aq) + NO3(aq) one mole of knog forms one mole of Nos" moles 4 NO, formed from 0.0960 mol KNO Imo NO3 = 0.0260 mol KNO, X . Tmal kitos -0.0260 x 1 mol No, = 0.0260 md Nos

page.:04 let the volume of Mg (NO) is x litres number & moles of Mg(NO3) = xL x 0.416 mol Il - 0,416x mol The equation for the dissociation of mg (NO3), is as follows 1. mg (NO3)2 - Mg ²+ + 2NO3 one mole of mg (NO3)2 forms of moles of Nog calculate the number of moles of Nos formed from 0.416 se mof Mg(NO3), is as follows number of moles of NO3 = 0,416x mol ng (NO3)2 x 2 mol Nos - inol Mg(NO3)2 0.416 * 2 mol NO +0.832 x mol No3 Total number of moles of No : 10.0260 +0.832x) volume of KNO₃ = 255. olx lt 1000 not = 255

page:05 = 0.055 L Total volume = (0.255 +XL Total concentration at No . (0.0260-40.832x) mo : 10.255 +x) L = 0.298.M (0,0260 +0,8384) -0.278 10.255+x) 0.0260 +0.839% = 0.248 (0.955+x) 0.0260 +0.839% = 0.07089+0.278x 0.832x -0.48% 0.0210-0.07089 - 0 0.5547-0.04489=0 0.554% = 0.04489 20.04489 D: 554 - = 0.0810 d volume = 0, 0810 < x 1000 m2 = 0.0810 X 1000 ML

page=06 = 81.0 ml Therefore, volume of Mg(NO3), should be added is 81.0 me (VII) osmotic pressure osmotic pressure - iMRT i = Van't Hof factor M = molasity solution R = univemsal gous constant -> 0.08206 Latmi molik. T = absolute temperature in k Van't Hoff factor of Nad - T = 47 fahrenheit Osmotic pressure Fax 0.50 x79 %0.08206 = 1 X 6. 31862 6. 31862 Therefore, Osmotic pressure = 6.31862