Question

The boiling point of water is 100.0°C at 1 atmosphere. How many grams of manganese(II) bromide (214.7 g/mol), must be dissolved in 270.0 grams of water to raise the boiling point by 0.300°C? Refer to the table for the necessary boiling or freezing point constant Solvent Formula Water H20 Ethanol CH3CH2OH Chloroform CHCI: Benzene CH Diethyl ether CH2CH OCH CH3 K (°C/m) Kr(°C/m) 0.512 1.86 1.22 1.99 3.67 2.53 5.12 2.02 g manganese(II) bromide.

Answer 1

The problem is based on the colligative property which is elevation in boiling point.

Data The in on the elevation given problem is based boiling point. is given ou follows: Ali = Kb. m .....(1). Here, o tb is eleuation in boiling point of pure solvent (Here water) by addition of solute C MoBr2 kb is molal elevation constant and m is the molellity of solcetion. eqnw is also given as follows: Atb = kb. W2 M, W, cann. (2) Here, Atb = 0.300°c. kb = 0.512 °C/m. w (weight of my Br = f M (molar mass of Motor = 214.7 g/mol. W, (weight of the water) = 270.0 g. A Tb = kb. N2 M₂. W .: 0.300 °C = 0.512 °C/ W2 &14g_'x 270.09 mol We = 0.300°C X 214.7 9 moi' X270.0 g 0.512 G mi molality having unit Here, m represents the mol per kg.

Date W Davolo - 0.300'c X 814.77 g.moT' X 210.09 0:512 Bloc .m = 0.300°C X 214.7.g.moi' X 270.09 - 0.512 °C. ( kgmol. kg.')" We = 0.300°Ć x 21414 x103 kap. moflx 270.0 X10369 0.512 od kg. mot Why = 0.033.9.66 kg Wy - 33.996 g of Moore Thus, grams of mo Br must be dissolved is 133.996 g