Which aqueous solution will have the lowest freezing point? (A) 0.040 m glucose (B) 0.025 m KBr (C) 0.020 m NaCl (D) 0.020 m BaCl_2

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more "im" value is is lowest freezing Point Here m= molality iz vant Hoff factor i = 1 + (n-1) a a = degree of Tonisation Here 100% Ionisation takes place <= 1 (tor electrolyte) n= number of Lons a 0.040m glucose m= 0.04om glucose is non electrolyte i= lt (n-1) * i= lt (n-1) o L1=1 im 1X0.040 m im 11 0.040 b 0.02sm KB8 m= 0.025 m KBO kt & Br

number of Ions n = 2 2 2 = 1 i = lt (n-1) * = 1 + (2-1) 1 ( 1+ It I i = 2. im = 2 x 0.025 im = 0.05 C 0.02om Nach m = 0.02om & 1 number of ions n=2 [nad Nattci lion lion أ i= lt (n-1) a = 1+(2-1) 1 i = 2 im 2 X 0.020 2x im = 0.04 0.020m Bacle Bach is more m= 0.020m x = 1 im value n=3 [Ba²+ + 2c1 3 3 +201) it is lowest lion 2ion freezing point i = ltcn-1) 2 1+(3-1) I correct option D - 3 3x0.0200.06 = = = im -

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