What is the molality of a solution of 75.2 g AgClO4 dissolved in 885 g C6H6?
Molality = moles of solute / kg of solvent solute = 44.9 grams of C10H8. There are approx 128 g/mole in C10H8. 44.9 grams x 1/128 moles/gram = 0.351 moles (approx) convert 175 grams C6H6 into 0.175 kilograms Molality = 0.351 moles solute (or C10H8) / 0.175 kg solvent (C6H6) Answer is 2.00 Molal
C6H6 is this a hydrocarbon
yes C6H6 IS Benzen.
Formula: C6H6
3g
hcl+c6h6
35g C6H6* 1mol C6H6/78.108g C6H6 * 1 mol C6H5Br/ 1 mol C6H6 * 157g C6H5Br/1mol C6H5Br = 70.35g of Bromobrezene.
C6h6
C6H6 would be good, as both CS2 and C6H6 are organic and non polar.
C6h6
CH
c6h6