Answer
molality =
moles of solute\divmass of solvent (kg)
a. mass of 1L soln = 1000mL×1.08g \div 1mL= 1080g
mass of water = 1080 g − (2.50 mol NaCl × (58.44 g NaCl \div 1mol NaCl)) = 934 g = 0.934 kg
m = 2.50mol NaCl \div 0.934 kg H2O = 2.68m
Work Step by Step
B. 100 g of the solution contains 48.2 g KBr and 51.8 g H2O.
mol of KBr = 48.2g KBr × (1mol KBr \div 119.0gKBr)= 0.405 mol KBr
mass of H2O (in kg) = 51.8g H2O× (1kg \div 1000g)= 0.0518kgH2O
m = 0.405 mol KBr\div 0.0518 kg H2O = 7.82m
