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To my knowledge, no. Ordinary first-order logic quantifies only over individuals (none of which are literally true) rather than over truth-valued things such as sentences or propositions. Thus there's nothing in first-order logic to which the predicate "is true" can apply. For that you need higher-order logic, which is a topic of controversy in its own right.

By "set of all true propositions," I take it you mean "a set of all the true propositions there are," i.e., the extension of the predicate "is a true proposition." A Cantorian argument due to Patrick Grim concludes that no such set is possible. It works by

reductio. Let T be any set containing all of the true propositions. If T exists, then it has infinitely many members, but that doesn't affect the argument. Now consider the power set of T -- P(T) -- which is the set whose members are all of thesubsetsof T. It's provable that any set has more subsets than it has members. With respect to each of those subsets in P(T), there is a true proposition concerning whether the propositionSnow is whitebelongs to that subset. It follows, then, that there are more true propositions than there are members of T, contrary to the assumption that T containsallthe true propositions there are. So no such set as T exists.