1.221×10^3 g of ethylene glycol, C2H4(OH)2, are dissolved in 1.1×10^3 g of water, H2O. What is the molality of ethylene glycol? A) 1.1 m
B) 1.9 m
C) 18 m
D) 6.8 × 10^4 m

The correct answer is:   bri hope this answer was ! : )

Hi + naoh -> h2o + nal

D=m/v =57.9/3 = 19.3 g/cm^3

The product formed when bao and h2o  react together is                        ba(oh)2that is  bao + h2o = ba(oh)21 mole of    bao    react  1 mole of h2o  to    form  1 mole  of ba(oh)2