suatu larutan 45% Massa NaNO3 mempunyai Kerapatan 1,368 gram/ml Tentukan: a. Fraksi mol NaNO3 dan XH20 Kemolalan NaNO3 b. kemolalan NaNO3 c. molaritas larutan NaNO3

a.Fraksi mol NaNO3 & H2O Diket : 45%NaNO3 ρ = 1,368 gr/ml Mr H2O = 18 Mr NaNO3 = 85 V air (asumsikan) = 100 ml >Mencari Massa Larutan,Massa NaNO3,Massa H2O mLarutan = ρ × V mH2O = mLarutan-mNaNO3 = 1,368 gr/ml × 100 ml = 136,8-61,56 = 136,8 gr = 75,24 gr mNaNO3 = % × mLarutan = 45% × 136,8 gr = 61,56 gr >Mencari mol Terlarut (nt) & mol pelarut (np) nt = m/mr np = m/mr = 61,56/85 = 75,24/18 = 0,724 mol = 4,18 mol >Mencari Fraksi Mol NaNO3 & H2O Xt = nt/nt+np Xp = np/np+nt = 0,724/0,724+4,18 = 4,18/4,18+0,724 = 0,147 = 0,852 >Mencari molalitas NaNO3 m NaNO3 = nt × 1000/gp air = 0,724 × 1000/75,24 = 9,622 m >Mencari Konsentrasi NaNO3 M NaNO3 = gtNaNO3/mr × 1000/V = 61,56/85 × 1000/100 = 7,24 M