How many moles and numbers of ions of each type are present in thefollowing aque

Question

How many moles and numbers of ions of each type are present in thefollowing aqueous solutions?
(a) 4.0 mL of 1.10 M magnesium chloride
moles of ions number of ions
Mg2+ mol ions
Cl - mol ions

(b) 259 mL of a solution containing 5.35 g/L aluminumsulfate
moles of ions number of ions
Al3+ mol ions
SO42- mol ions

(c) 2.27 L of a solution containing 5.61 1021 formula units oflithium nitrate per liter
moles of ions number of ions
Li+ mol ions
NO3- mol ions

Explanation / Answer

How many moles and numbers of ions of each type are present in thefollowing aqueous solutions? (a) 4.0 mL of 1.10 M magnesium chloride moles of ions number of ions Mg2+ mol ions Cl - mol ions moles of MgCl2 = 4 mL * (1 L /1000 mL) *(1.10 mol MgCl2/L) = 0.0044 mol MgCl2 =>   0.0044 molMg2+          and 0.0044 mol*(6.022e23 molecules/ mol) = 2.64968e21 = 2.65 x1021 molecules =>   2*0.0044 = 0.0088 molCl-                 =>   5.29936e21 molecules ~ 5.30 x 1021molecules (b) 259 mL of a solution containing 5.35 g/L aluminumsulfate moles of ions number of ions Al3+ mol ions SO42- mol ions 0.259 L * (5.35 g Al2(SO4)3 / L) * (1 mol /342.15 g) = 0.00404983195 mol Al2(SO4)3 => moles Al = 2*0.00404983195 = 0.0080996639 molesAl3+   = 0.00810 moles Al3+ => 0.00810 mol Al3+ (6.022e23 molecules/mol) =4.8776176e21 = 4.88 x 1021 molecules Al3+ => moles SO42- = 3*(0.00404983195) =0.0121494959 = 0.0121 mol SO42- => 0.0121 mol *(6.022e23 molecules/mol) = 7.31642643e21 = 7.32 x1021 molecules SO42- (c) 2.27 L of a solution containing 5.61 1021 formula units oflithium nitrate per liter moles of ions number of ions Li+ mol ions NO3- mol ions note sure what "5.61 1021 formula units" is. Are there 5.611021formulat units of 5.61*1021 formula units. Seems it is thelatter => 5.61*1021= 5727.81 formula units 2.27 L *(5727.81 formula units/ L)= 13002.1287 formula units ofLiNO3 LiNO3 => Li+    +   NO3- => 1.30 x 104 ion Li+ =>   1.30e4 formulat units * ( 1mol /6.022e23 formulatunits) = 2.15910473e-20 = 2.16 x 10-20 moles Li+ B/c of 1:1 stoichiometry the same number of ions of Li+ and NO3- are formed. Same formoles

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