66

Oct. 18, '49
M.W. = 181.9
Weight of V(subscript)2 O (subscript)5 = 0.185g
dissolved in dilute HCl
Filtered out the residue by means of a filter paper
of 0.897 gr
Filtrate evaporated to 9 C.C. giving a H(superscript)1 - signal
equivalent to h(subscript)2O + 0.002M - MnSO(subscript)4

Weight of filter paper + residue = 0.900g
Weight of residue = 0.003g
Weight of V(subscript)2O(subscript)5 in 9 c.c. sol (superscript)n = 0.182g
Molarity of V in sol (superscript)n = 2x0.182/0/182 x9 = 0.222//